WILEY
BENZIGER
GEOMETRY
UNDER THE EDITORIAL DIRECTION OF
ROY DUBISCH AND ISABELLE P. RUCKER
LAWRENCE A. RINGENBERG
RICHARD S. PRESSER
Benziger, Inc. New York, Beverly Hills
'',n Association v/itn
John Wiley & Sons, Inc. New York, London, Sydney, Toronto
EDITORS
ROY DUBISCH, Professor of Mathematics at the University of Washington, has
been writing for both teachers and students for many years. He has been Associate
Editor of the American Mathematical Monthly and Editor of the Mathematics Maga
zine. He has several books to his credit including The Teaching of Mathematics
(Wiley, 1963) and is the author of numerous journal articles. Since 1961 Profes
sor Dubisoh has served on the Mathematics Steering Committee of the African Edu
cation Project and has participated in several workshops and institutes in various
African countries. He has also directed and lectured at several NSF summer in
stitutes. He is a past Vice President of the National Council of Teachers of
Mathematics and has served on the Teacher Training Panel of the Committee on the
Undergraduate Program in Mathematics, the Advisory Board and Panel on Supple
mentary Publications of the School Mathematics Study Group, and the Board of
Governors of the Mathematical Association of America. He has also been active in
many other professional organizations,
ISABELLE P. RUCKER, Supervisor of Mathematics of the State Board of Education
of Virginia, is a former teacher of both elementary and secondary school mathe
matics. She has participated in summer mathematics institutes and an NSF
Academic Year Institute at he University of Virginia and has published numerous
articles and book reviews in professional journals. In addition to being active in
other professional organizations, Mrs. Rucker has served on the Advisory Board,
the Executive Committee, and the Panel on Supplementary Publications of the
School Mathematics Study Group, For the National Council of Teachers of Math
ematics, she has served as Chairman of the Committee on Plans and Proposals
and as a member of the Committee on Affiliated Groups, She was Program Chair
man for the Golden Jubilee Annual Meeting in 1970 and Is an active member of
the Committee on Meetings of the Council. In addition, she has held offices in the
Association of State Supervisors of Mathematics,
Copyright © 1973 by John Witty and $om, Inc.
All nghu r*t»rv*d, PubJfthctf «lmullan»t>imy )n Canoda.
No port of ihl» book may be r»pr«doe»d by any mtoni, nor
frflnimi««d, nor troniloted inlo a moctoln* lannyoflfl
wifhoul Iho wriltfln p«rmin1on of (h» pvblUhir,
ISBN 0471 P4549.8
Printed In ihe UnFfsd Sratoi of Afntrka
10 967*34321
AUTHORS
RICHARD 5. PRESSER, Coordinator of Secondary School Mathematics of the
Michigan City Schools, Michigan City, Indiana, is a former writer for the Minne
sota National Laboratory, the School Mathematics Study Group, and the African
Education Program of Educational Services, Inc. Mr, Presser has many years of
teaching experience at the secondary school level and is active In the Indiana
Council of Teachers of Mathematics and the Central Association of Science and
Mathematics Teachers.
LAWRENCE A. RINGENBERG, Professor of Mathematics and Dean. College
of Letters and Science, Eastern Illinois University, Charleston, Illinois, was Head
of the Department of Mathematics from 1947 to J 967. He is a former writer for
the School Mathematics Study Group, the author of College Geometry (Wiley,
1968) and Informal Geometry (Wiley, 1967), and is currentry serving as a mem
ber of the editorial board and one of the authors of the forthcoming NCTM Year
book on geometry. Dean Ringenberg was Coordinator of the U.S.A.LD. Mathe
matics and Physics Courses Program in East Pakistan for Intermediate College
teachers in East Pakistan during the summer of 1966. Since 1960 he has been
a member of the Charleston, Illinois. Unit District Board of Education,
Preface
Geometry is one of a series of five mathematics textbooks for
junior and senior high schools. It is designed for a oneyear course
for students with a background of informal geometry and elementary
algebra such as that included in Mathematics J, Mathematics II, and
Algebra of this series.
In its formal development of Euclidean geometry, this textbook
features an integrated treatment of plane and solid geometry with
an early introduction of coordinates. Coordinates on a line, in a
plane, and in space relate numbers ami points. They are used to
gain knowledge of geometrical figures and to simplify the develop
ment of formal geometry,
Students use their knowledge of elementary algebra throughout
the book to help them learn geometry. In doing this they maintain
and strengthen their competence in algebra.
The postulates in this book form the basis of a rigorous, yet
plausible, development of a first course in formal Euclidean geometry.
In some instances, statements that are proved in more advanced
treatments are accepted as postulates here. This has been done to
decrease the length of the development and Lo make the develop
ment appropriate for high school students.
Geomctrv is an important subject because it is practical and
useful and at the same lime abstract and theoretical. There arc two
main objectives in this geometry textbook. One is to help students
learn a body of important facts about geometrical figures. These
facts, interpret* all v. are facts about the space in which we
will Preface
live. These facts are important for intelligent citizenship and for suc
cess in many careers. The other main objective is to help students
attain a degree of mathematical maturity. In the elementary and
junior high schools, students are encouraged to learn by observing
and manipulating physical objects. There is considerable emphasis
On intuitive and inductive reasoning. The power and beauty of
mathematics, however, is due primarily to its abstractness. The
generality of its theorems makes possible a variety of applications.
Understanding mathematics "*in die abstract" is tantamount to
understanding the deductive method in mathematics. In studying
this book students will develop their capacity to reason deductively
and hence their ability to read and write proofs.
Contents
Chapter 1
Points, Lines, and Planes 1
I.l Informal Geometry 1 / 1.2 The Idea of Formal
Geometry 4 / 1.3 The Ideas of Point, line, and Plane
10 / I A Sets 14 / 1.5 Conjunctions and Disjunctions
18 / 1.6 The Incidence Relationships of Points and Lines
22 / 1.7 The Incidence Relationships of Points, T.incs, and
Planes 28 / Chapter Summary 34 / Review Exercises 35
Chapter 2
Separation and Related Concepts 40
2.1 Introduction 41 / 2.2 The Betweenness Postulates
41 / 2.3 Using Betweenness to Make Definitions 47 / 2.4
The Concept of an Angle 52 / 2.5 The Separation Postulates
56 / 2.6 Interiors and Exteriors of Angles 67 / 2.7
Triangles and Quadrilaterals 72 / 2.8 Properties of Equality
and Number Operations 80 / 2*9 Solving Equations
S3 / 2.10 Equivalent Equations 86 / Chapter Summary
92 / Review Exercises 93
Chapter 3
Distance and Coordinate Systems 96
3.1 Introduction 97 / 3.2 Distance 98 / 3.3 Line
Coordinate Systems 105 / 3.4 Rays, Segments, and
Coordinates 107 / 3.5 Segments and Congruence
Contents
111 / 3.6 Two Coordinate Systems on a Line 116 / 3.7
Points of Division 129 / Chapter Summary 135 / Review
Exercises 135
Chapter 4
Angles, RayCoordinates, and Polygons 138
4.1 Introduction 139 / 4.2 Angle Measure and
Congruence 140 / 4.3 Betweenness for Rays 146 / 4.4
RayCoordinates and the Protractor Postulate 152 / 4*5
Some Properties of Angles 160 / 4,6 Interiors of Angles
168 / 4.7 Adjacent Angles and Perpendicularity
170 / 4,8 Polygons 177 / 4,9 Dihedral Angles
181 / Chapter Sunn nary 184 / Review Exercises 185
Chapter 5
Congruence of Triangles 188
5.1 Introduction 189 / 5.2 Congruence of Triangles
190 / 5.3 "IfThcn" Statements and Their Converses
197 / 5.4 The Use of Conditional Statements in Proofs
200 / 5.5 Proofs in TwoColumn Form 203 / 5M The
Congruence Postulates for Triangles 208 / 5.7 Using the
S.A.S., A.S.A.. and S.S.S. Postulates in Wilting Proofs
216 , 5.8 Isosceles Triangles 227 / 5.9 Medians and
Perpendicular Bisectors 234 / Chapter Summary 242 /
Review Exercises 244
Chapter 6
Inequalities in Triangles 246
6.1 Introduction 247 / 0,2 Inequalities for Numbers
248 / 6.3 The Exterior Angle Theorem 254/6.4
Inequalities Involving Triangles 262 / Chapter Summary
274 / Review^ Exercises 275
Chapter 7
Parallelism 280
7.1 Introduction 281 / 7.2 Definitions 282 / 7.3
Existence of Parallel Lines 284 / 7.4 Transversals and
Associated Angles 286 / 7.5 Some Parallel Line Theorems
Contents xl
/ 7.0 The Parallel Postulate and Some Theorems
301 / 7.7 Parallelism for Segments; Parallelograms
308/7.8 Parallelism for Rays 314 / 7.9 Some Theorems
on Triangles and Quadrilaterals 318 / Chapter Summary
325 / Review Exercises 325
Chapter 8
Perpendicularity and Parallelism in Space 328
8.1 Introduction 329 / 8.2 A Peipendieularity Definition
332 / &3 A Basic Perpendicularity Theorem ,334 / 8.4
Other Perpendicularity Theorems 333 / 8.5 Parallelism for
Lines and Planes 347 / 8.6 Parallelism and Perpendicularity
352 / Chapter Summary 359 / Review Exercises 380
Chapter 9
Area and the Pythagorean Theorem 362
9.1 Introduction 363 / 9,2 Area Ideas 364 / 9.4 Area
Postulates 368 / 9.4 Area Formulas 372 / 9.5
Pythagorean Theorem 379 / Chapter Summary 385 /
Review Exercises 386
Chapter 10
Similarity 390
10.1 Introduction 391 / 10.2 Proportionality 392 / 10.3
Properties of Proportionalities 396 / J 0.4 Similarities
Between Polygons 401 / 10.5 Some Iength Proportionalities
409 / 10,6 Triangle Similarity Theorems 415 / 10.7
Similarities in Right Triangles 422 / 10.8 Some Right Triangle
Theorems 429 / Chapter Summary 436 / Review
Exercises 437
Chapter U
Coordinates in a Plane 440
11.1 Introduction 441 / 1 1.2 A Coordinate System in a
Plane 442 / 1 L3 Graphs in a Plane 447 / 11.4 Distance
Formulas 452 / 11.5 The Midpoint Formula 458 / 11.6
Parametric linear Equations 464 / 11.7 Slope 473 / 1 1.8
Other Equations of Lines 481 / 11.9 Proofs Using Coordinates
494 / Chapter Summary 506 / Review Exercises 507
xii Contents
Chapter 12
Coordinates in Space 510
12.1 A Coordinate System in Space 511 / 12.2 A Distance
Formula 519 / 12.3 Parametric Equations for a line in
Space 523 / 12.4 Equations of Planes 526 / 12.5
Symmetric Equations for a Line 539 / Chapter Summary
544 / Review Exercises 544
Chapter 13
Circles and Spheres 546
13.1 Introduction 547 / 13.2 Circles and Spheres: Basic
Definitions 547 / 13.3 Tangent Lines 544 / 13.4 Tangent
Pknes 567 / 13.5 Circular Arcs, Arc Measure 576 / 13.6
Intercepted Arcs, Inscribed Angles, Angle Measure
584 / 13.7 Segment*: of Chords, Tangents, and Secants
596 / Chapter Summary 607 / Review Exercises 608
Chapter 14
Circumferences and Areas of Circles 612
14.1 Introduction 613 / 14.2 Polygons 614 / 14.3
Regular Polygons and Circles 622 / 14,4 The Circumference
of a Circle 634 / 14.5 Areas of Circles; Arc Length; Sector
of a Circle 643 / Chapter Summary 636 / Review
Exercises 658
Chapter 15
Areas and Volumes of Solids 662
15 .1 Introduction 663 / 15.2 Prisms 664 / 15.3
Pyramids 672 / 15.4 Areas and Volumes of Prisms and
Cylinders 680 / 15.5 Volumes of Pyramids and Cones
688 / 15.6 Surface Areas of Spheres and Volumes of
Spherical Regions 697 / Chapter Summary 702 / Review
Exercises 704
Contents xiii
Appendix Al
list of Symbols A 1
List of Postulates A2
List of Definitions A5
List of Theorems A20
Table of Squares, Cubes, Square Roots, Cube Roots A42
Glossary G]
Index 11
Chapter
Sam Folk /Mankmeycr
Points,
Lines, and
Planes
1.1 INFORMAL GEOMETRY
Geometry began informally more than 2(KX) years ago in Babylon
and Egypt. Tlie meaning of geometry is, literally, "earth measure
ment"; hence it is not surprising thai ruilv knowledge of the subject
was concerned largely with measurements of lengths, areas, and vol
umes. Such knowledge was a necessity, especially to the aneient Egyp»
tians who almost annually were forced to restore to their riverside
farms the boundary markers which were washed away by heavy flood
ing of the River Nile.
Then, as well as now, men learned by experience. The ancient
Egyptians developed many rulesofthmnb for surveying fields and
roads and for making calculations related to building dwellings and
pyramids. These rules were based on many observations, intuition, and
reasoning. As long ago as 500 B*G men knew how to find the areas of
rectangles, triangles, and trapezoids, but they had made little progress
in formal geometry, that is, in developing geometry as a system that
explains why the rules worked.
Points, Lines, and Piano*
Chapter l
As a school subject today geometry is bolh informal and formal
In the elementary schools geometry is largely informal. It is physical
geometry. Students work with physical objects or with pictures thai
represent physical objects, General statement 1 ? or rules are based on
intuitive reasoning and inductive reasoning. Intuitive reasoning is what
might be called common sense, or, as some might say, reasoning in a
hurry. Inductive reasoning is reasoning based on numerous examples.
Wc discuss an example of each.
Intuitive Reasoning
Let us take for granted that we know what is meant by a triangle.
Figure 1! shows a right triangle A ABC with right angle at C
figure M
Suppose that the lengths of the sides AO and BC are b inches and
a inches, respectively. What b the area of the triangle? We say "area
of the triangje/' but we really mean the area of the figure or region
that consists of the triangle and its interior. In earlier mathematics
classes we have learned that the area is \ab square inches. Suppose
someone asks why. One answer might be: *"It is intuitively obvious.
You can see it by looking at a picture of a rectangle with sides of lengths
a and h and with one diagonal drawn." (See Figure 12.) If the person
who asked why responds with "Now I see why," he is responding to
an intuitive feeling of the "tightness" of tilings rather than to a logical
argument or to inductive reasoning. This is an example of intuitive
reasoning.
Figure 12
1.1 Informal Geometry
Inductive Reasoning
Ixl us take for granted that we know what is meant by an angle
and its measure. We record the measures of angles as numbers, omit
ting the degree symbol. This is consistent with our formal point of view
regarding measure developed later. Suppose that each student in an
elementary geometry class is given a set of plastic triangles numl Hired
1, 2, 3, 4, as suggested by Figure 13, and a protractor. The instructions
direct the students to find and record the measures of the angles of the
triangles and then to Hnd the sum of the measures for each triangle as
in the following table. The object of the lesson is to make it plausible
that the sum of the measures of the angles of a triangle is 180, If the
students reason that this is a correct conclusion on the basis of the
measurements they have made, this is an example of inductive
reasoning.
mLA
mlB
m t C
Triangle 1
90
32
58
180
2
75
26
78
179
3
95
25
62
182
4
65
25
90
180
Points, Lines, and Planes : i ,ip.er
1.2 THE IDEA OF FORMAL GEOMETRY
Suppose that as a result of inductive reasoning we htive made a
general statement about all triangles, or about all rectangles, or about
all circles. We think that the statement is true, but how can we know
for certain? Is it possible to know something for certain about all tri
angles? It would seem necessary to test every triangle, which is clearlv
impossible. Therefore a different approach is required, the formal
approach.
The formal approach involves deductive reasoning, that is, logical
arguments by which general statements are obtained from previously
accepted statements Two examples follow.
Deductive Reasoning
Example Suppose that you are given a set of plastic convex quadri
laterals numbered 1, 2, 3 t . . . , 10 as suggested by Figure i4 Suppose
that there are really ten of them, although only four of them are shown.
Figure m
Suppose also that you know that the sum of the measures of t lie angles
of every triangle Is 1 80. Perhaps you know this through the example
of inductive reasoning in Section LL At any rate you know or are told
this. This knowledge about all triangles is considered as "given" in the
situation of this example. The instructions are to find the sum of the
measures of the angles of each quadrilateral without making any meas
urements using a protractor.
A quadrilateral is convex only if it has the property that if either
pair of its opposite vertices is joined by a segment, called a diagonal
of the quadrilateral, then all of that diagonal except its endpoints lies
inside the quadrilateral. Figure I 5a shows a convex quadrilateral. In
Figure 1 5b the segment PH, except for the points P and R, lies outside
the quadrilateral. Hence PQRS is not a convex quadrilateral.
A picture of a quadrilateral and one of its diagonals provides a clue
for finding the sum of the measures of the angles, (See Figure 16.)
Quadrilateral ABCD may be considered representative of any convex
quadrilateral. From this flgtirc we see that the sum of the measures of
1,2 The Idea of Formal Geometry 5
(a) Convex
(b) Not convex
Figure 15
Figure 16
the angles of a quadrilateral is the sum of the measures of the angles
of two triangles. Therefore the sum of the measures of the angles of a
quadrilateral is 2 * 180 or 360 Without using a protractor we decide
that the sum of the measures of the angles of each quadrilateral is 360.
We arrive at this conclusion by deductive reasoning.
Let us examine this example carefully so that we may understand
more clearly the nature of deductive reasoning. We took some things
for granted or as given. From these we deduced the answer, or con
clusion. The given statements are called hypotheses, or collectively,
the hypothesis. Thus the essence of deductive reasoning is to obtain a
conclusion from a hypothesis The hypothesis may or may not be true.
If it is true in a given situation and if the reasoning involved in reaching
die conclusion is correct, then the conclusion is true in that situation.
Let us look at this idea expressed in symbols. Suppose that wc are
given the hypothesis, denoted by If, and we want to establish the con
clusion, denoted by C* What we want to prove is not "C" but "If ff,
then C." Many statements in mathematics are in this "If H t then C"
form How can such a statement be useful? It is useful in any situation
where the "H " is true. Because if we know "H" is true and if we know
"If IT, then C " is true, we also know that "C" is true. In the problem
about quadrilaterals, the hypothesis and the conclusion may be identi
fied as follows;
H: ABCD is a convex quadrilateral.
C: The sum of the measures of the angles of ABCD is 360.
6 Prints, Lines, and Planes
If ABCD is a quadrilateral like the one in Figure T 5a, then '*H " is
true, and since "If H, then C" is true, it is correct to conclude that "C"
is true. If ABCD is a quadrilateral like the one in Figure l5b, then
"H" is false, and although "If ti % then C " is true, it is incorrect to con
clude that "C " is true.
Example This example in which deductive reasoning is used to es
tablish a general statement concerns a property of the natural numbers,
1, 2, 3, . . . . Undoubtedly, you know an even number is a number that
is 2 times an integer. Thus x is an even number if there is an integer ij
such that x = 2y. Some even numbers are 2, 16, 168, and 2466. If you
add any two of these numbers, you get an even number. Is it then true
that the sum of any two even numbers is an even number? Of course
it is. Let us prove it, however, by using deductive reasoning.
H: x and y are even numbers.
C: x + y is an even number.
Our task is to prove: If H, then C.
Proof: Since x is even, there is an integer u such that x = 2n. Since
y is even, there is an integer c such that y s 2c. Then
x + \j = 2u + 2c
and, using the distributive property, we get
x + y = 2{u + v).
But u +■ 15 is an integer. Therefore, since x \yis2 times an integer,
x + y is an even number.
Notice that we did not prove that x is even or that y is even. We
proved that if x and y are oven numbers, then *f y is an even number.
As we said, formal geometry involves deductive reasoning. An im
portant feature of a formal geometry is its structure or arrangement.
Hie geometry of this book is elementary Euclidean geometry, carefully
arranged so that we can see how the various parts fit together and how
some things depend on other things. Formal geometry might be
thought of as geometry for the lazy person. In formal geometry we
prefer a general statement that tells something significant about all
triangles rather than a hundred statements about a hundred triangles.
Our starting point in formal geometry is a set of statements about
some of the simplest objects of geometry. We do not try to tell what
these objects are by definitions. Definitions would involve other words
that we would need to define in turn, and so we accept some concepts
1,2 The Idea of Formal Geometry
as basic and undefined. We might decide, for example, to start with
triangles because almost everyone lias an idea of what a triangle is. But
do we actually know what a triangle is? A triangle is made up of three
segments. The notion of segment is more basic than the notion of tri
angle. Every segment is a set of points. The idea of a point is more
fundamental than the idea of a segment. Every segment is a part of
some line. Perhaps a line is simpler to think about than a segment
In formal geometry we usually consider points, lines, and planes
as the basic building blocks. We do not define these words. How then
can we be sure that wc know anytlung at all about them? On the basis
of our experience with physical objects we identify the most basic
properties that points, lines, and planes have in relation to one an
other. We formulate these as statements that we accept without proof.
We call these statements postulates.
The foundation for formal geometry, then, is a set of statements,
the postulates, which we accept without proof. "The postulates are
statements about the basic objects in geometry. We agree that what
we know about these objects is what the postulates say, and nothing
more, at least at the start. What else can we possibly know about these
objects? We can know what %ve formally assume in the definitions and
what we deduce by logical reasoning.
As an example suppose that we start with the following six postu
lates and one definition. (This is not for keeps— just for this example!)
Our notations arc selfexplanatory.
POST U L ATE A Every line i g a set of p oints and contains at least
two distinct points.
POSTULATE B Every segment is a subset of a line and contains
(besides other points perhaps) exactly two distinct points called its
endpoints.
POSTULATE C If A and B are any two distinct points, there is
exactly one line AB containing A and B, and exactly one segment AB
with A and B as endpoints.
POSTULATE D Every plane is a set of points and contains at
least three points that do not all he on one line, that is, three noncol
linear points.
POSTULATE E If A, B, C arc three noncollinear points, then
there is exactly one plane containing A, B, and C.
POSTULATE F If a plane contains two distinct points A and B,
then it contains the segment AB.
S Points, Unes, and Planes
Chapter 1
Definition If A t B t C arc any three noncolHnear points, then
the onion of the three segments AB, BE, and ^A is a triangle;
we denote it by A ABC.
In this example, what do we know for sure? We know what is said
in these seven statements, the six postulates and the definition. What
else do we know? We know anything else that we deduce by logical
reasoning from them. We shall deduce one statement and call it a
theorem.
THEOREM Every plane contains at least one triangle.
Proof; Let a (read "alpha") be any plane. Then a contains three non
colltiiear points (Postulate D)j call them A, fl, C. Then there is exactly
one segment AB with endpoints A and B t exactly one segment BC with
endpoints B and Q and exactly one segment Cli with endpoints C and
A (Postulate C). Then there is a triangle AABC (by our definition).
Each of the segments AB, B£, CA lies m a {Postulate F). Therefore
AABC, which is their union, lies in a and the proof is complete.
In proving a theorem it is usually a good idea to include one or more
inures l.i> suggest the given .situation. In this Mliialirn we start witli 8
plane a as suggested in Figure 17, Next we reason deductively to get
duee noncollincar points A, B, C in a as suggested in Figure 18,
Finally, we reason deductively to show that there is a triangle AABC
and that it lies in a as suggested in Figure 19*
Figure 17
Our theorem may seem trivial to you. Our goal, however, was sim
ply to show an easy example of a theorem obtained by deductive rea
soning from a set of postulates.
1,2 The Idea of Formal Geomttry
EXERCISES 1.2
In Exercises 10, a situation is given and a question is asked Obtain an an
swer in each case using intuitive reasoning,
1. Given ihc two numbers
x a 21.3 + 27.4 and y = 28.5 + 16.2 + 5.9,
which is larger, x or y?
2. Given the two numbers
p = (1.6X754) and q = (L6X89G),
is pq greater than, equal to, or less than (2)(76SX896)?
3. The following figure shows a square and a parallelogram with side
lengths unci angle measures as labeled. Do these figures have equal
areas? If not, which one has the larger area?
4 The figure below shows two parallel lines. Is there a plane containing
these two lines?
«
5. The figure shows a rectangular box with the vertices labeled. Is there a
plane containing the four points ft, G, C* A?
6. Given the situation of Exercise 5, is there a plane containing the four
points A, R, C t H?
7.
H:
C;
8. H:
C:
9,
H:
C:
10.
H.
C:
TL
H:
n.
12.
H.
C:
10 Points, Lines, and Planes Chapter 1
■ In Exercises 712, a hypothesis H and a conclusion C are given. State
whether you think the conclusion follows logically from the hypothesis. Be
prepared to defend your answer.
x is an odd number.
x 2 is mi odd number.
ar is a multiple of 3.
a 2 is a multiple of 6.
X is a multiple of 3,
x 2 Is a multiple of 9.
A ABC is a right triangle with sides of lengths 3 in. 3 4 in. ; and 5 in.
The area of &ABC is 6 sq in.
p and q are two distinct intersecting planes.
The intersection of p and <] is a line.
S is a sphere and p is a plane that intersects 55.
The intersection of S and p is a circle or a point.
13 THE IDEAS OF POINT, LINE, AND PLANE
Every day in the world around us we observe objects of different
sizes and shapes. We notice that some of these objects have corners,
edges, and sides, and that some of their parts are "straight," some are
"flat," and some are "round." Touching and seeing certain objects help
ns to classify them according to their characteristics.
Tn arithmetic the idea of a number is a mathematical idea that grew
out of a need to classify certain sets according to how many objects
they contained. But. no one has ever seen or touched a number. In ge
ometry the ideas of point, line, and plane arc mathematical ideas that
grew out of a need to classify certain sets of figures and to measure thei r
boundaries or the regions bounded by the figures. But no one has ever
seen or touched a point, a line, or a plane. Just as in arithmetic you
studied numbers and the operations on them, in geometry you will
study points, lines, and planes and how they relate to one another.
In the same way that deductive reasoning must be based on certain
assumptions (postulates) that we accept without proof so must our
definitions be based on certain terms that we make no attempt to de
fine. Tliis is necessary in order to avoid "circular" definitions, that is,
a chain of definitions which eventually comes back to the first word
being defined. If a person does not know the meaning of any of the
words in the chain, the definition is of no value to him. For example,
in defining the word "dimension" one dictionary uses the word "mag
nitude." It defines "magnitude" In terms of "size." When looking up
1.3 Th» Idtai of Point, Line, and Plane 1 1
"size" in this same dictionary, it gives "dimension or magnitude,"
Thus, if someone does not know the meaning of any of the terms mag
nitude, size, or dimension, the dictionary is not very useful. Somewhere
in the cycle it is necessary to know the meaning of a word based on
experience.
The basic undefined terms in our geometry are p>trrf T line, and
plane. Our postulates give these terms the meaning that we wish them
to have. You undoubtedly already have an intuitive feeling for the con
cept of a point, a line, or a plane. We can think in a vague fashion of a
point as having "position" but no "size"; of a line as being "straight;"
having "direction," but no "width"; and a plane as being "Hat" but
having no "thickness." When we "mark a point" or "draw a line" on
our paper or on the chalkboard, we are merely drawing a picture of
what we think a point or a line should be, These pictures or figures help
us to see and discover some of the relationships that exist among points,
lines, and planes and help us to keep these relationships straight in our
minds. However, any deductions that we reach must be justified
strictly on the basis of our postulates, definitions, and theorems and not
on what appears to be true from a figure. As we progress in our study
of geometry, we will use figures more and more freely. If our figures
are drawn carefully enough, they generally « ill not give us false infor
mation or lead us to false conclusions.
Indeed, trying to deduce al the theorems and work all the prob
lems in this text without drawing any figures would be a tedious and
difficult task. It would be somewhat like a carpenter attempting to
construct a house from memory, without the aid of any blueprints or
floor plans. The resulting structure might be quite different from the
house he planned to build.
We use figures frequently to help explain what our postulates, defi
nitions, and theorems say. You are cneouruged to do the same. When
it is practical, you should restate a theorem or a problem in terms of a
figure that shows the relationships that arc given in the theorem. If you
are careful not to include in the figure any special properties that are
not given in the theorem, then the figure should be a valuable aid in
understanding the theorem and also in proving it.
EXKRCISES 1.3
1. Name three physical objects that convey the idea of a point.
2. Name three physical objects that convey the idea of a line.
3. Name three physical objects that convey the idea of a plane.
12 Points, Lines, and Planes
Chapter 1
Exercises 49 refer lo the cube shown in Figure 110. Answer each question
in Exercises 47 with the word point, lino, plane, or space,
4. Each cornei' (vertex) of the cube suggests
ft!®.
5. Each side (face) of the cube suggests a [7.
6. Each edge of the cube (such as ATS) suggests
7. The interior of the cube is a subset of [Tj.
8* How many vertices Joes the cube have? How
many faces? How many edges?
&. Let V represent the number of vertices, E
the number of edges, and F the number of
laces. Compute V — E + F for the cube.
Figure MO
In Exercises 1017, copy and complete the table on the following page for
the pyramids and prisms shown below.
14
15.
1,3 The Ideas of Point, Line, and Plane 13
17,
i«ZI>
E
10.
E + F
12.
13.
14,
15.
10.
17.
18. I5id you get the same number for V — £ + Fin Exercises 917? If you
have counted correctly, V — E ■+■ F equals 2 in each case. Figures like
those in Exercises 1CM7 are called polyhedrons. On the basis of your an
swers for Exercises 917, do you think that V — E + F = 2 for every
polyhedron? Compute V T jE, F, and V — E + Ffor the starshaped poly
hedron shown in Figure 111, (You can construct this polyhedron by
gluing together 16 triangles and 2 squares.)
Figure MI
14 Points, Lines, and Planes Chapter 1
1.4 SETS
In our development of formal geometry we shall often speak of
sets of points, or sets of lines, or sets of planes, as well as sets of num
bers. Indeed, we shall agree later in this chapter that every line is a set
of points and that every plane is a set of points. Although you are fa
miliar with the language of sets, we s h all review it briefly.
Recall that a set is simply a collection of objects. The objects may
be numbers, people, points, or whatever. The objects of u set are called
the members or the elements of a set. The symbol € indicates that
an object is an clement of a given set. Thus, to indicate that the num
ber 2 is an clcmenL of the set {1, 2, 3, 4), we write
2 €{1,2, 3, 4}.
Wc read the expression 2 £ {1, 2, 3, 4} as "2 is an element of the set
{ 1, 2, 3 t 4}" or simply as "2 is in { 1, 2, 3, 4}/' Similarly, if point A is an
element of the set of points in line J, we write
A£l
and read this as "'point A Is on line h"
It is often convenient to use setbuilder notation to indicate the
members of a particular set For example,
{x : x is an integer and —2 < x < 4}
is just another way of naming the set { — 1, 0, 1, 2, 3}, We read the ex
pression {x : x is an integer and —2 < x < 4} as "the set of all num
bers x such that X is an integer and X k greater than —2 and less than
4." If we wished to include —2 and 4 in this set, we would write
— 2 < x < 4 rather than —2 < x < 4.
Example 1 Let R represent the set of all real numbers. Suppose thai
wc wish to picture (graph) the set
C = {x : x € H and 3 < x < 3}
on a number line. Another way of writing the expression —3 < x <[ 3
is x ^> — 3 and x < 3.
The number line in Figure 112 pictures the set
A = (s : x € R and x > 3}.
■o— ! i
6 5 4 321 1 2 3 4
Figure 112
1.4 Sets 15
Note that the point corresponding to the number —3 has been circled
to indicate that —3 is not an element of this set.
The number line in Figure 113 pictures the set
B = {x : x £ R and x ^ 3}.
Why is the point corresponding to the number 3 circled and shaded in
the figure?
B
«
■S 54321 I 2 3 4 5 fi
Figure 11*3
It is clear from the two number lines in Figures 112 and 113 that
the numbers that are common to both sets (those that belong to both
A and R) are the real numbers between —3 and 3 and including 3.
Tills is shown on a single number line in Figure 114.
c
4 o ♦ — ! ►
654321 J 2 3 4 5 6
Figure 114
We call the set shown in Figure 114 the intersection of the two
sets in Figure 113. The intersection symbol D is used in forming a
symbol to denote the intersection of two sets. Thus to indicate the in
tersection of the two sets
A = {x : x £ H and x > 3}
and
B = {x : x £ R and x £ 3}
we write A n B and we see that tills intersection is the set
C = {x : x £ R and 3 < x <, 3}.
rherefore C = A P B.
In set language the connective "and" is related to intersection.
"Thus the elements that belong to sets A and B in Example 1 are those
elements that are common to the two sets; in other words, the inter
section of the two sets is the set C.
It may happen that the intersection of two sets is empty, meaimi Li
that the two sets have no elements in common. The symbol indicates
the empty or null set For example, since the set E of even integers and
the set O of odd integers have no integers in common we write
En 0=0.
In formal geometry, when we say that one set intersects another
set, we mean that the two sets have at least one element ra common.
16 Points, Lines, and Planes Chapter 1
In this case, the intersection of the two sets cannot be the empty set.
For example, consider the three sets
D = {2, 3. 5, 7},
F = {0, 4, 6, 9},
G = (3, 5, 8, 10}.
Sets D and F do not intersect since they have no elements in common,
and we write DOF= , However, sets D and G do intersect and
we write DrC=(3,5).
Tn listing the elements of a set, as for D, F, and G, the order in which
the elements appear is not important. Thus, if U = {2, 3, 5, 7} } then
we may also write D = (2, 5, 7, 3}, The intersection of two sets is in
dependent of order, too. Thus, if A and B are any two sets, then
a n n = b n a.
Example 2 Let set = {x : x € J* and x < 3 or * > 3}, Let us
picture the sets
A = {*:x€Rand*< 3}
and
i?={x:x€fiandx>3}
separately on two parallel number lines. (Note: A and B are not the
same sets as in Example 1.)
The number line (a) in Figure 115 shows the graph of the set
A = {x : *€ Rand*< 3}
and the numt>er line (b) shows the graph of the set
8= {x :x€Randx>3}.
A
(a) 4
0>)
_
i
6
ft
*
3
2
1
i
2
3
4 a
u
4
B
►
6 5 4 S 2 t
Figure 115
Those numbers that belong to set A or set B are the numbers that he
to the left of 3 or to the right of 3. Figure 116 on page 17 shows this
on a single number line.
1.4 SMs 17
D
4 i . >
6 54321 1 a U i 5 6
Figure I 10
We call the set shown in Figure 116 llic union of the two sets in
Figiire 115. The union symbol U is used in forming a symbol to denote
the union of two sets. Thus to indicate 'the union of two sets
A = {x : x £ R and x < 3}
and
B = {% : % € R mix > 3} f
we write A U B and see that this union is the set
D= {x : *EHand*< 3ori>3).
Therefore D = A U B.
Example 2 illustrates how the connective "or" is related to unions
in the language of sets. Thus the elements that belong to sets A or B
are those elements that are cither in set A or in set B or in both. The
set of all elements that belong to A Of B is the union of the two sets
A and B which, we have seen, is the set D.
The union of two sets is independent of their order. Hence, if
A and B are any sets, then A J B = B J A,
The ideas of union and intersection are extended easily to apply
to any number of sets. For example, if A, B, and C are sets, then
A n B n C is the set of all elements each of which is in aU three of
the sets A. B, and C Describe the intersection of the two sets A and B
of Example 2L
Example 3 Let A = { l t 2, 3, 4, 5} and B = (3, 4, 5, 6, 7}. The union
of A and B is the set of numbers in A or B, and we write
AU B= {.1, 2, 3* 4, 5, 6, 7}.
It should be noted that if a number is in set A, or in set B, or in both
set A and set B, then this number is in A J B. Thus 1 and 2 belong
in A U B because 1 and 2 are in set A. The numbers 6 and 7 belong
in A U B because 6 and 7 are in set B. Finally, the numbers 3, 4, and 5
belong in A J B because 3 T 4, and 5 are in both set A and set B, The
intersection of A and B is the set whose elements are common to both
sels A and B, and we write
AH B= {3,4.5}.
18 Points, Lines, and Planes Chapter 1
Recall that a set A is a subset of set B if every element of A is also
an element of B. (Write A C B for "A is a subset of B") Thus, if
A = {«, b), then each of the sets («}, [h], {a, b} t and is a Subset
of A.
EXERCISES 1.4
L Let A = {x : x£ flandx< lJandB = {x : x£ R andx > I).
(a) Graph the set A. the set B, and the set A (1 B on separate number
lines,
(h) Use setbuilder notation to indicate the intersection of sets A and H.
(c) Use setbuilder notation to indicate the union of sets A and B.
Kxercises 28 refer to the sets
A= {3, 1, 1,3},B= (2.. 0,2, 4}, and
C={2, 1,0,1,2}.
2* Find A n B. A Find A U CL
3. Find A n C. 7. Find A UBUC
4. Find BnC. 8. Find (AtBJfl C.
5. Find A U B,
9. The figure below shows the graph of a set of numbers on a real number
hue. Use setbuilder notation to describe this set.
7 6 5 4 3 2 1
>
10, The figure below shows the graph of a set of numbers on a real number
line. Use setbuilder notation to describe this set
"
6 5 4 3 a 1
1.5 CONJUNCTIONS AND DISJUNCTIONS
Similar to the relationships of two sets and their union and inter
section are the relationships of two statements and their conjunction
and disjunction. In mathematics it is very important to understand
clearly the relationship of truth in given statements to truth in state
ments formed from the given ones. In this section we consider the rela
tion of two statements to the statements formed from them by con
necting them first by "and" and then by "or."
1.5 Conjunctions and Disjunctions 19
Suppose that p is a statement and that q is a statement. (By a state*
ment we mean a sentence that is cither true or false.) We call the state
ment "p and (/"the conjunction of the statement p and the statement
q> For example, if p is the statement
4 = 4
and q is the statement
4 = 5 t
the conjunction of p and q is the statement
4 = 4 and 4 = 5.
We agree to call a conjunction of two statements true if and only if the
statements are both true.
► If p is true and q is true, then p and q is true; if p and q is tru e,
then p is true and q is true.
Thus the conjunction 4 = 4 and 4 = 5 is false because the state
ment 4=5 is false. On the other hand, the conjunction 4=1 unci
5 = 5 is true. Of course, if both of the statements p and q are false,
then the conjunction p and q is false.
Let us see how the conjunction of two statements is related to the
intersection of two sets. In Example 1 of Section 1.4, we considered
the intersection of two sets
A = {« i x £ R and x > 3} and B = {x : x £ R and x < 3}.
We saw that the intersection of these two sets is the set of aE numbers
that belong to both A and B. In other words, if X is a number, then it is
true that x is an element of A and of 8 (x £ A D B) if and only if it is
true that x is an element of A and it is true that x is an element of B.
For example, 2 £ A and 2 £ B; hence 2 £ A P B. On the other hand,
5 £ A and 5 £ B (5 is not an element of B), hence 5 £ A n J3.
Thus if the statements i£A and x £ B are true, then the conjunc
tion of these statements (x £ A n B) is also true. But if either of the
statements x £A and * £ B is false, then * € A n J? is false.
Now consider the statement p or q where p and q are statements.
We call the statement p or q the disjunction of the two statements p
and C/. For example, if £? is the statement
4 = 4
and f/ is the statement
4 = 5,
the disjunction of p and q is the statement 4 = 4 or 4 = 5.
20
Points. Lines, and Planes
Chapter I
We agree to call the disjunction of two statements true if and only if
either, or both, of Ihc two statements is true,
► If /> is true and q is true, then p or q is true; if p is true and q
is false, then povq is true; if /> is false and q is true, then p or
q is true; if p or q is true, then either p is true and q is false, or
p is false and q is true, or p and q are both true.
Thus the disjunction 4 = 4 or 4 = 5 is considered to be true be
cause 4 = 4 is true. Similarly, the disfunction 4 as 4 or 5 s 5 is t me
since the statements 4 = 4 and 5 = 5 are both true. On the other
hand, the disjunction 4 = 5 or 7 < 5 is False since I Kith of the state
ments 4 = 5 and 7 < 5 are false.
Let us see how the disjunction of two statements is related to the
union of two sets. In Example 2 of Section 1.4 we considered the union
of the two sets
A = (x : x € B and x < 3} and B = {x : x £ R and * > 3}.
We saw that the union of these two sets is the set of numbers belong
ing to either A or B. In other wortfc. if x is a number, then it is true that
i is an element of A or B (x € A U B) if and only if at least one of the
statements x is an element of A and x is an element of B is true. For
example, 5 € A, but 5 € B; hence 5 C A U R On the other hand,
2 g A and 2 £ B; hence 2 t A U B,
Thus, if at least one of the statements x £ A and x £ B is true, then
the disjunction of these two statements (x £ A U B) is also true. But
if both of the statements x € A and x £ B are false, then the statement
x 6 A U B is false.
FAERCISES 1.5
Exercises 16 refer to Figure J 17. Imagine that each of the four lines a, b,
c.din the figure is a set of points.
FJeuru I 17
3 .5 Conjunctions and Disjunctions 21
L The set {£} is the intersection of which two lines?
2. What is the intersection of h and c?
3* What is the Intersection of the three sets a t h, and rf?
4 Does « fl fc = a P fi?
5, What is a H c?
6. What is the intersection of the four sets a, b, c, and d?
The following table shows the possible truth values (T for true, F for false)
that can be assigned to the statements p and q. In Exercises 710, copy and
complete the table for the statements p and q and p or q.
p and q p or q
7.
T
T
T
8.
T
F
F
ft.
F
T
CD
10.
F
F
DQ
11. If x £ A and x £ B t is ran element of A H B? Of A U B?
In Exercises 1215. P and Q represent sets. Copy and complete the table
using T for true or F for false.
*€ P
x€Q
x 6 P n Q
XC p \J Q
12.
T
T
T
7]
13.
T
F
a
m
14.
F
T
3
3
15.
F
F
13
F
16. Two sets arc said Lo be equal if they contain exactly the same elements.
If A n {x ; ar is an integer and  1 < % < 5} and B  {0, 1, 2 S 3, 4),
does A = B? Why or why not? If C = {2. 3, 0, 1 , 4 } } does R = C?
17, In Exercise 1.6 is A a subset of B? Is B u subset of A? Is 5 a subset of C?
Is C a subset of B?
IS. If A C B and B C A are both true, what can you conclude?
19. If A and B arc sets and A = B i does B = A?
20. If A and B arc sets and A C B, is B C A?
22 Points. Lines, and Planet Chapter 1
21. (a) List all the subsets of the set { 1, 2, 3}. (Remember that the empty
set is a subset of every set.)
(b) How many distinct subsets does a set co nsisting of three elements
have? Four elements;? n elements?
22. If A = and & = {3, 4, 5, 6}, what is A H B? A J F?
23. (a) Given that x is a number such that x 2 = 16, find one possible value
for x. Is there another possible value for x? What is the solution set
of the equation x 2 = 16?
(b) Write die solution set of the equation x 2 + 7 = 56.
■ In Exercises 2429, write the solution set of the equation.
24. 3a  5 = 13 27. 2(* + 3) = 5a  8
25. 5(2*  9) = 10 28. x*  5 = 31
26. 3(2*  7) = t + 9 29. x 2  2x  15 =
30. If F = f (x, y) : x £ H, tj C R. and y = 2x  3}, we find that when
i=2 T y = 2'2 — 3 = 1. Thus the ordered pair (2, 1} is an element
of the set F. To check this, note that "2 £ R, 1 € R, and 1 = 2 ■ 2  3"
is a true statement. This is the statement you get if you replace x by 2
and y by 1 m the sentence which follows the colon in the setbuilder
symbol
(a) What value of tj is paired with x when x = ^?
(b) Find five more ordered pairs of numbers that belong to the set F.
31. Describe the solution set of the equation 2(x — 2) = 2x — 4.
32. Describe the solution set of the equation 2{x — 2) = 2x  5.
1.6 THE INCIDENCE RELATIONSHIPS
OF POINTS AND LINES
We are now ready to begin the formal development of geometry.
We do not state all the postulates of our formal geometry at once, but
rather introduce them throughout the text as needed. In this chapter.
we state the first eight of que postulates. These postulates are con
cerned with points on lines, in planes, and in space; lines through
points, in planes, in space ; and planes through points, through lines,
and in space. Figure 118 shows a picture of a line I and a point F. As
the figure suggests, point F is on line / and line / passes through point F.
Since t is a set of points and P is an element of this set, it is correct to say
that F is "in" I or that F is a point "of" L However, we usually say that
Figure 1
1,6 Relationships of Points and Lirv« 23
Pis "on" I and that 1 "passes through" P, Another way of saying this is
to say that the point and line are incident, it is for this reason that our
first postulates, which we state in this section and in Section 1.7, are
called Incidence Postulates:
For our first postulates we draw on our experience with physical
representations of points and lines. When you use your ruler or
straightedge to draw a line through two points, such as P and Q in Fig
ure 119, you are, of course, actually working with physical representa
tions of points and lines and not the abstract points and lines of our
geometry.
Figure 119
We decide what we are going to accept without proof about geo
metric points and lines by looking at the real world. In fact, we need
only look at Figure 119 to arrive at our first three postulates. First,
however, wc state two definitions.
Definition 1.1 Space is the set of all points
Definition 1.2 The points of a set are collinear if and only
if there is a line which contains all of them. The points of a
set are n uncoil in ear if and only if there is no line which con
tains all of them.
Plane Postulates of Incidence
POSTULATE 1 (The TiireePoint Postulate) Space contains
at least throe noncollinear points.
POSTULATE 2 (The LinePoint Postulate) Every line is a set
of points and contains at least two distinct points.
POSTULATE 3 (The PointLine Postulate) For every two
distinct points, there is one and only one line mat contains both points.
24 Points. Lin«. and Planes Chapter 1
Postulate 3 is sometimes shortened to read "two points determine
a line." When we use the word "determine" here, we use it in the
sense of Postulate 3, which states that, given two distinct points, there
is exactly one line which contains them. Similarly, when we say that
"three noncolliiicar points determine three lines," we mean that there
are exacdy three lines each containing two of the three points.
You may feel that these postulates do not tell us very much about
points and lines. Your experience has led you to believe that there arc
a great many points on a line and in space, and you may wonder why
we do not say so in the postulates. We feel that it is more instructive
to proceed as we have done, Shortly we will have introduced enough
postulates that we will be able to prove that there are infinitely many
points on a line and in space. However, there is not much we can prove
from only the three postulates. Figure 119 should suggest to you at
least two statements concerning lines that are not stilted in the postu
lates. Before formulating these statements, we will introduce some
Symbols for points, lines, and planes.
Notation. Capital (upper case) letters are used to denote points and
small (lower case) letters to denote lines. For example, in Figure 1 19
P, Q, and R denote points and /, m, and n denote lines. We may also
name a line by naming two points that determine the line. Thus cither
of the symbols PQ or Qr could be used to name the line that is deter
mined by points P and Q. It may also be convenient to name several
different lines by using the same letter with distinguishing subscripts
such as h, I2, fa* * • • «
In the figures we represent a line by the symbol 4 ►
where the arrowheads are used to indicate that the line does not stop
where our picture stops but continues indefinitely in both directions.
We often use the Creek letters a, /?, v, . , , (alpha, beta, gamma, . . .)
to name a plane, Sometimes it is convenient to name a plane by nam
ing three noncollinear points that the plane contains, such as "plane
FQfi* in Figure 119.
Let us return now to our discussion of the relationships between
points and lines. Figure 119 suggests, but our postulates do not tell
us, that there are at least three distinct lines in space. Let us now at
tempt to deduce this statement from the three postulates we already
have.
The Three Point Postulate tells us that there are at least three
points in space that do not all lie on one line. For convenience we name
these points P t Q> and R. It follows from the PointLine Postulate that
1.6 Relationships of Fbints and Lines 25
points P and Q determine exactly one line PQ, that points Q and R de
terminc exactly one line QR, and that points P and R determine exacdy
one line PR. No two of these lines arc the same. For if they were, points
P, Q, and R would be collinear, and this would contradict the fact that
P, {), R do not all lie on one line. This completes the proof of the fol
lowing theorem. (Usually the proof of a theorem follows the statement
of the theorem. For Theorem 1.1, however, the proof precedes the
statement of the theorem.)
THEOREM 1, 1 There are at least three distinct lines in space.
We observe from Figure 119 that no two distinct lines intersect
in more than one point We proceed now to deduce this statement
from our postulates.
Tf /, and t 2 are any two distinct lines that intersect, then, by defini
tion of intersect, we know they have at least one point in common.
Call this point P. Now, either they have a second point in common or
they do noL Let us suppose that they do have a second point Q in com
mon as suggested In Figure 120.
Figure 120
The PointLine Postulate tells us that for every two distinct points
there is one and only one line that contains them. Thus h = h (diat is,
they are the same line). Rut this contradicts our hypothesis that k and
h are distinct tines. Since our supposition thai l 1 and fe intersect in a
second point leads us to a contradiction, we must conclude that h and
I2 intersect in no more than one point. We have proved the following
theorem.
THEOREM 1,2 If two distinct lines intersect, then they intersect
in exactly one point.
Does Figure 120 confuse you because the marks that represent
lines do not look as you think straight lines should look? Actually the
word straight is not part of our formal geometry. We want a formal
26 Points. Lines, and Plans* Chapter 1
geometry of points, lines, and planes that is consistent with our experi
ences in informal geometry. In particular, we want lines in our formal
geometry to have the property of straight ness. It is true that we cannot
draw a convincing picture of two distinct lines with two distinct points
in common. Tills is evidence at the informal geometry level consistent
with Theorem 1.2.
EXERCISES 1.8
L Winch postulate asserts that there are at least two distinct points on
everj' line?
2. Restate the LinePoint Postulate and the PointLine Postulate in your
own words.
3. Restate the definition of "intersect" as the word applies to two sets of
points.
4* If points A, B, C, and D are distinct, and if line / contains points A, B,
and C and line in contains points A, C, and D t what can you conclude
from this?
5. In proving Theorem 1 . 1 wc started with three noncollincar points and
showed that there are three distinct lines each containing two of the
three noncollincar points. Suppose we start with four points, no three
of them collincar. How many distinct lines are there each containing
two of the four points?
6. Suppose we .start with five distinct points, no three of them coBmear.
How many distinct lines are there each containing two of the five
points?
7. Suppose we start with six distinct points, no three of them collincar.
How many distinct lines are there each containing two of the six
points?
8. Consider your answers to Exercises 5, 6*, and 7. Can you predict how
many lines there are for seven distinct points each containing two of
the seven points if no three of the points are collinear? If so, how
many?
9. Let m and n lie different lines. Let A be a point snch that A g m and
A <E n, Let B he a point such tliat B £ m and B C «• What must be true
of points A and B? What postulate or theorem supports your answer?
10. Let A and B be different points. Let line i% contain A and B and let line
/a contain A and B, What can you conclude iihout / t and l z ? What pos
tulate or theorem supports your conclusion?
1.6 R#l»tiofl ships of Points and Lines 27
11. Consider the set consisting of the five points,, and no others, named in
the following figure. If the points appear to be eollinear, assume that
they are,
(a) Identify, by listing the members, two subsets each containing three
eollinear points.
(b) Identify, by listing the members, four subsets each containing four
noneollmear paints of which three points are unilinear.
(c) Ust the members of one subset contai ning four noneollinear points
of which no three are eollinear.
(d) Name three lines which are not drawn in the figure, but each of
which contains two points named in the sketch.
12. From which postulate does it follow that no line contains all points of
space?
challenge pnonLUM* Start with three distinct objects such as three boys
named Jerry, Jim, and John, Create a structure as follows: Each boy is a
point and there is no other point. Thus space = {Jerry, Jim, John}. There
are three distinct lines a, h, and c (and no others) as follows: a = {jerry,
Jim}, b = {Jim. John), c = (Jerry, John}. Exercises 1330 are questions
about the Jerry JimJohn structure. Kxercises 3134 are general questions.
13* Are the points, Jerry, Jim, and John, eollinear?
14. Does space contain at least three noncollinear points?
15. Does the structure satisfy the ThreePoint Postulate?
16. Is a a set of points?
17. Does a contain at least two distinct points?
18. Is b a set of points?
19. Does h contain at least two distinct points?
20. Is c a set of points?
21. Docs c contain at least two distinct points?
22. Does the structure satisfy the Line Point Postulate?
23. Is {Jerry, J tin} a set of two distinct points?
28 Points, Lines, and Planes Chapter 1
24, Is {Jerry, John) a set of two distinct points?
25. Is {Jim, John} a set of two distinct points?
26. Are there other sets of two distinct points?
27, Is there one and only one line that contains Jim and John?
2S, Is there one and only one line that contains John and Jerry?
29, Is there one and only one line that contains Jerry and Jim?
JO. Does the structure satisfy the Point Line Postulate?
31, Can we prove, using only the plane postulates of incidence, that space
contains at least four distinct points?
32, What do the Plane Postulates of Incidence tell us about the nature of
an individual point?
33, What do the first three postulates tell ns about the straightness of a line?
34, What do the first three postulates tell us about segments?
1.7 THE INCIDENCE RELATIONSHIPS
OF POINTS, LINES, AND PLANES
We are now ready to list our remaining Incidence Postulates. We
begin by stating the following useful definition.
Definition 1.3 The points of a set arc eoplanar if and only
if there is a plane %vhich contains all of them. The points of a
set are noncxmluiuir if and only if there is no plane which con
tains all of them.
Space Postulates of Incidence
POSTULATE 4 (The FourPoint Postulate) Space contains at
least four nonco planar points.
POSTULATE 5 (The PlanePoint Postulate) Every plane is a
set of points and contains at least three noncollinear points.
POSTULATE C (The PointPlane Postulate) For every set of
three noncollinear points there is one and only one plane that concur
them,
POSTULATE 7 (The FlatPlane Postulate) If two distinct
points of a line belong to a plane, then every point of the line belongs
to that plane.
POSTULATE 8 (The PlaneIntersection Postulate) If two dis
tinct planes intersect, then their intersection Is a line.
1,7 Relationships of Points, Lines, and Planes 29
Note that our postulates do not include a statement about the num
ber of planes in space. Theorem 1.3 states that there are at least two
distinct planes in space. In the Exercises that follow you will be asked
to prove that there are at least four distinct planes in space.
From Postulate J we know that there arc aL least three noneollincar
points in space. Call these points A, B, and C, By Postulate 6 there is
exactly one plane a that contains these points. From Postulate 4 there
is a fourth point D that is not in plane a. Points A, B t and D are not
unilinear, for if they were, point D would lie in plane a. Why? By
Postulate 6 again there is exactly one plane /? that contains points A,
B, and D. Planes a and f$ are not the same plane, for if they were, points
A, B, C. and D would be coplanar, which is contrary to the way they
were chosen. We have proved the following theorem.
THEORF.M L3 Space contains at least two distinct planes.
It is often helpful to draw diagrams or pictures showing the rela
tionships between points, lines, and planes when making; a deduction
such as the one for Theorem 1,3. Figure 121 shows points A, B, and C
in plane a and points A, B. and D in plane /?. (Although we often use
a quadrilateral to represent a plane, be careful not to think of the sides
of the quadrilateral as the "edges" or "ends" of the plane.)
Figure 12] also helps us to "see" that when two planes intersect,
their intersection is a line as we have stated in Postulate 8. What is the
intersection of a line and a plane not containing the line? When we
look at Figure 122, it appears that the intersection is a single point
L
Figure 121
1
Figure 14S
30 Points, Linti, and Planet Chapter 1
THEOREM 1.4 if a line intersects a plane thai does not contain
the line, then the intersection is a single point.
We can prove Theorem 1 ,4 by the same technique used in proving
Theorem 1.2. Head again the proof of Theorem 1.2 which appears
hefore the statement of the theorem. Now try to answer the questions
in the argument (proof) that follows*
By hypothesis and the definition of intersect, line m and plane of
(Figure 122} have at least one point F in common. How do we know
that plane a does not contain every point of line m? Suppose line mi
and plane a have a second point Q in common. (Here we are examining
one of only two possibilities: either the line and plane have a second
point in common or they do not. Of course, we are trying to prove
that they do not.) But if a plane contains two distinct points of a line,
then it contains the entire Line. (What postulate are we using here?)
What does this last conclusion, namely, that the plane contains the
entire line, contradict? Since we have agreed to accept our postulates
as true statements al>out points, lines, and planes, what must we con
clude? Docs this prove the theorem?
It follows from Postulate 6 (the PointPlane Postulate) that three
noncoUinear points determine exactly one plane. That is, there is
exactly one plane which contains any set of three noncoUinear points.
We conclude this section with two additional theorems regarding
the existence of planes.
THEOREM 1.5 If m is a line and P is a point not on m, then there
is exactly one plane that contains m and P.
Note that we must prove two tilings.
1. There is at least one plane t hat contains m and P.
& There is no more than one plane that contains m and P.
These two statements illustrate the ideas of existence and uniqueness.
When we prove existence, we show that there is at least one plane con
taining m and P. When we prove uniqueness, we show that there is
at most one such plane. If we can prove both existence and unique
ness, then we know that there is exactly one such plane.
In the following proof name the postulate or theorem that justifies
the statement preceding the question "Why?".
Proof of I. Existence. There is a plane a that contains line m and
point P (Figure 123).
1.7 Relationships of Points, lines, and Planes 3 1
There arc at least two distinct points on line m. Why? Call these
points A and B. The points A and B do not lie on any line except m.
Why? By hypothesis, point P does not lie on line m. Hence points A,
B, and V arc noncoUinear. There is a plane tt Lhat contains points A,
B t and P. Why? If plane a contains points A and B, then it contains
line ro. Why? Hence there is a plane that contains line m and point P,
Mgurc 123
Proof of 2. Uniqueness. There is no more than one plane tJiii
contains line m and point P. Suppose that there is a second plane /?
which contains m and P. Then /? contains the noncoUinear points A, B t
and P. Thus we have two distinct planes* a and j8 s each containing
three noncoUinear points. What postulate does this contradict? Does
litis prove that there is no more than one plane containing m and F?
THEOREM L6 If two distinct lines intersect, then there is
exactly one plane that contains them.
Let lines m and n intersect in the point P (Figure 124).
a
Figure 124
We must prove the following:
1. Etafatencft There is at least one plane that contains Hi and m.
2. Uniqueness. There is no more than one plane that contains
m and it,
32 Points, Lints, and Planes Chapter 1
In the following proofs name the postulate or theorem that justifies
the statement preceding the question "Why?"
Proof of 1; Lines m and n have exactly one point in common. Why?
There is a point A on m that is different from P t Why? A is not on n by
Theorem 1.2. (Two distinct lines intersect in no more than one point.)
There is a plane a which contains A and n. Why? Plane a contains F
and A and therefore contains m. Why? Hence there is a plane a which
contains m and n.
Proof of 2; Suppose there is a second plane that contains both m
and n. Then plane contains A because A is on m. Thus both a and fi
contain A and n. What theorem docs this last statement contradict?
Does this prove that there is no more than one plane a containing m
and n?
EXERCISES 1.7
Refer to Figure 125 in working Exercise* 112. Assume that points A t B,
and C are in the plane of the paper and that point D is not
Figure 119
1* Name all the planes, such as ABC, that are determined by the points
in the figure
2. Name atl the lines that are determined by the points in the figure. One
of them is line A£.
3. How many lines arc there in each plane?
4. How many planes contain line BD? Name them. Is this number of
planes the same for every line?
5. How many lines are on every point? Name the lines that are on point D.
6. How many points do plane ABC and line 55 have in common? What
theorem justifies your conclusion?
1.7 Relationships of Points, Lines, and Planes 33
7. Name two planes, unci name a point that is contained in both planes.
Which postulate tells ils that these two planes have a line in common?
Name the line.
8. There are at least how many points common to every pair of planes in
the figure?
9 Two planes which do not intersect arc said to be parallel. On the basis
of our first eight postulates, do you think it can be proved that there
must be two parallel planes in space?
10, Is it possible for three planes to have one and only one point in com
mon? If your answer is "Yes/' name three planes and the single point
they have in common.
11, Two noncoplanar lines in space are said to be skew lines. Name three
pairs of lines in the figure such that each pair is a pair of skew lines. Can
a pair of skew lines intersect? Why?
12. Can three lines have a single point in common and be noncoplanar
lines?
13, Let P be a point in plane a. On the basis of only our first eight postu
lates do you think that we can conclude that there are three distinct
Hues in plane a that contain point F? Give reasons for your answer.
14. How many points do the planes of the ceiling, front wall, and side wall
of your classroom have in common? On the basis of only our first eight
postulates do you think it can be proved tint there arc three distinct
planes that have more than one point in common?
15. Using three pages of your book as physical examples of planes, what
would you guess to be the intersection of three distinct planes if then
intersection contained more tlian one point?
16\ I Ay the edge of your ruler along the top of your desk. Do all the points
on the edge of the ruler seem to touch the desk? Which of our postulates
docs this illustrate?
In Exercises 1725, write a short paragraph to show how our postulates or
theorems can be used to prove the .statement.
17. Every line is contained in at least one piano.
18. There are at least three distinct lines in every plane.
19. There are at least four distinct planes in space.
20. There are at least six distinct lines in space.
21. If four points are noncoplanar, then they are noneollinear.
22. If four points are noncoplanar, then any three of these points are
noneollinear.
23. There are at least three distinct lines through every point.
24. Every line is contained in at least two distinct planes.
25. Every point is contained in at least three distinct planes.
34 Points, Ones, and Planes Chapter i
26. challenge problem. Suppose, for this one exercise, that we replace
our Postulate 8 with this postulate: "If two distinct planes have one
point in common, then they have a second point in common." Prove
our replaced Postulate 8 as a theorem.
27. challenge problem. Suppose, just lor Exercises 27 and 28, that
space consists of four distinct points A, B, C, D and no others. Suppose
that there is one line a containing all four of the points A, B, C, D and
that there are no other linos. Suppose that there are no planes. Does
this structure satisfy Postulate 4? Since A, B, C, and D are not contained
m any plane, it follows that they arc noncoplanar, and hence Postulate
4 is satisfied.
Does this structure satisfy Postulate 5? Yes, it does. To show lliis
we must check each plane in the structure to see if it contains three
noncollinear points. If there is no plane that violates the property of
Postulate 5, then Postulate 5 is satisfied. Since there is no plane in the
structure, it follows that there is no plane that violates Postulate 5, and
hence Postulate 5 is satisfied. Show that Postulates 6, 7, and 8 arc satis
fied by this structure,
28. challenge p itonLEM . Does the structure of Exercise 2 7 satisfy Postu
late 1? Postulate 2? Postulate 3? Your intuition might tell you that Pos
tulate 4 includes Postulate 1 in the sense that Postulate 1 could he
proved once Posulate 4 is assumed. The structure of Exercise 27 pro
vides an answer to the question; Is it possible to satisfy Postulate 4, in
deed all of the postulate, from 2 through 8, without satisfying Postulate
1? In other words, is Postulate 1 independent of Postulates 2 through 8?
Why?
CHAPTER SUMMARY
In tliis first chapter we compared informal geometry and formal ge
ometry noting that geometry began informally many years ago as a collec
tion of rules to solve practical problems related to "earth measurement."
Although geometry as a school subject is both formal and informal, the ap
proach in tins book is predominantly formal.
Tile formal approach features carefully stated postulates, definitions,
and theorems. Proving theorems involves deductive reasoning. We dis
cussed examples of intuitive reasoning, inductive reasoning, and deductive
reasoning.
We reviewed the language of sets. If A and B are sets ? the INTERSEC
TION of A and B, denoted by A n B t is the set of elements contained both
in A and in B, The UNION of A and £, denoted by A U B, is the set of ele
ments contained in A or B, The intersection of two sets may be die
EM PTY or NULL SET denoted by . If two sets INTERSECT, they must
have at least one element in common.
Review Etercis** 35
If p and q arc statements; the CONJUNCTION of these two state
ments is the statement p and q. The conjunction of p and o; is true if both p
and ^ arc true, but false in all other cases. The DISJUNCTION of the two
statements is the statement porq. The disjunction of p and q is false if both
p and q are false, but true in all other cases*
In order to avoid circular definitions w« accepted POINT, LINE, and
PLANE as UNDEFINED TERMS in our formal geometry.
We defined SPACE to be the set of all points. The points of a set are
COLLINEAR if and only if there is a line that contains all of them. The
points of a set are COPLANAR if and only if there is a plane that contains
all of them.
Our first eight postulates are called INCIDENCE postulates and arc
listed below by name only. Be sure that you can state each of them in your
own words.
1. THE THREEPOINT POSTULATE
2. THE LINKPOINT POSTULATE
3. THE POINTLINE POSTULATE
4. THE FOURPOINT POSTULATE
5. THE PLANEPOINT POSTULATE
6. THE POINTPLANE POSTULATE
7. THE FLATPLANE POSTULATE
8. THE PLAxNEINTEftSECnON POSTULATE
In addition to these eight postulates we stated and proved six theorems.
Read these theorems again and study their proofs.
REVIEW EXERCISES
In Section L I an example involving the area of a right triangle was given to
illustrate intuition or iutuitive reasoning. In Exercises 110, state whether
or not you think the given example is a good example of intuitive reasoning.
If you think the decision made is false, indicate it.
1. Examine a rectangular box and decide that the opposite faces of a rec
tangular solid have the same size and .shape.
2. Examine a ball and decide that a plane through the center of a spherical
region (a Sphere and its interior) would separate it into two hemispheri
cal regions with equal volumes.
3. Examine a ball and decide that the area of a sphere is four times the
area of a circular cross section formed by a plane passing through the
center of the sphere.
4. Examine three distinct points on a line and decide that if A, B y C are any
three distinct points of aline with B between A and C, then the distance
from A to H is less than the distance from A to C.
36 Points, Unas, and Pfants
Chapter 1
5. Think about a triangle Trying in a plane a. Think about a line in or that
passes through a point F lying inside of T. Decide that this line must
contain two distinct points of T.
6*. Think about a triangle, a quadri
lateral, and a pentagon. Decide that
the sum of the measures of the
angles is the same for each.
7. Think about three books of 300
pages each standing upright on a
shelf as suggested in the figure. De
cide that it is farther from page 1 of
book I to page 300 of hook 1 II than
it is from page 300 of book 1 to page
1 of book III.
8, Think about a plane figure consisting of a triangle and the bisector of
one of its angles as suggested in the following figure. Decide that the
bisector of an angle of a triangle divides the opposite side into two seg
ments of equal length.
9. Tliink about a circle and a line that separates it into two arcs of equal
length. Decide that the line passes through the center of the circle,
10. Draw a triangle Mark one point on each side as close to the midpoint
as you can without measuring. These points are vertices of a second
triangle. Decide thai the area of the original triangle is four times the
area of the second triangle.
In Section LI an example involving the sum of the measures of the angles
of a triangle was given to Illustrate inductive reasoning. In Exercises 1120
draw one or more figures and make some measurements if you wish. Then
draw an appropriate conclusion.
11. Is the following statement true or false? The bisectors of the angles of
a triangle are concurrent, thai is, they all pass through the same point.
12. What is the sum of the measures of tlie angles of a pentagon? (Consider
only pentagons whose diagonals pass through the interior of the
pentagons.)
13. If A BCD is a quadrilateral with AB = BC, h it necessarily true that
CD = DA?
Review Exercises 37
14. If ABCD is a quadrilateral with AC = HD t is it necessarily true that
AB = BC?
15. If the sides of a quadrilateral have equal lengths, is it necessarily true
that its diagonals have equal lengths?
16. If the diagonals of a quadrilateral bisect each other, is the quadrilateral
necessarily a parallelogram?
17. If the sides of AABC have equal lengths, if D is the midpoint of BC,
If E is on line DA, if D is between A and E» and if AD — DE, is it true
that S?is parallel to AC?
18. If AABC is any triangle and if D is the midpoint of EC* is it necessarily
true that AABD and A A CD have equal areas?
19. If AABC is a triangle and if D is the midpoint of B~C, is it true that
AD = %AB + AQ?
20. If A ABC" is a scalene triangle (no two of its sides have the same length),
is it true that no two of its angles have equal measures?
Section 1,2 contains two examples of deductive reasoning. In Exercises 21
25, state whether you think the conclusion C (Ci, Ca, . .  , if more than one)
follows logically from the hypothesis H (Hi, 91% if more than one). Be
prepared to explain what assumptions you are making, if any, in addition
to those that are stated as hypotheses,
21. H. x = o
C: 2* f 3 = 13
22. Hi 2x + 3 = 13
Ct x = 5
23. Hi'. Every polygon is a plane figure,
ll'i P is a polygon.
& P is a plane figure.
24. B Let AABC and ADEF be given. If ml A = mLD, mlB =
B»Z£, and AB = DE, then AC = DFsad BC = EF.
G; tx;t AABD be given with C between A and D. If m / ABC =
ml VBC and mlACB = m I DCB, then AC = CD
Q: Let AABC and ADEF be given. If ml A ~ mZD, AB =
DE f and AC = DE, then BC = EF.
25. Hi: ATI cows eat grass.
H& Bessy is not a cow.
Ct Bessy does not eat grass,
26. If A is the set of even integers and B is the set of odd integers, describe
the union of A and B< the intersection of A and B. Do sets A and B
Intersect?
38 Points, Lines, and Plane* Chapter 1
■ Exercises 2738 refer to the three sets
Q= {3, 0, 3 r G),
<J= {3, 0, 3, 6,9},
T = [x \ x is an Integer,  3 <; x < 9, and x is divisible by 3),
27. Is Q C S? 33. Is T C S?
28. Is S C 9? 34. Docs S = 7?
29. Is Q c 7? 35. Find Q D S.
30. Is T C 0? 3fl  Uoe* D S = Q?
31. Does Q = T? 37. Find Q U S,
32. Is S c T? 38. Does £ U S = S?
39. If A = {x ; x £ R and x < 7} and B = (x : x £ /I and x > 3}, draw
the graphs of A, B, and A H Bon three separate number lines.
40. Describe the set A H /J in Exercise 39 using setbuilder notation.
41. If S = {x : x £ R and x > 5} and T = {x : x C R and x < 0), draw
the graphs of S, T, and S U Ton three separate number lines.
42. Describe the set S U T in Exercise 41 using setbuilder notation.
■ If p is the statement 7 = 7, q is the statement \/9 as 5, and r is the statement
— 5 < —2, determine in Exercises 4349 if the given "compound" state
ment is true or false.
43. p and q 47. q and r
44. p or q 48. q or r
45. n and r 49. (p and q) or r
46. por r
50, State the Flat Plane Postulate in your own words.
51. The following figure is a picture of two distinct "curves" that intersect
in two distinct points. State a postulate or theorem that would settle an
argument about whether these " curves'" might be lines in our formal
geometry.
52. Iff and Q are distinct points and m and n are lines, and if P £ m, Q € ™>
P E ?t, and Q £ n, what conclusion can you draw? What postulate or
theorem justifies this conclusion?
53. If m and n ai'e distinct lines and P and Q are points, and if m fl n = F
and m n n — Q, what can you conclude? What postulate or theorem
justifies this conclusion?
Review Exercises 39
54. Which postulate assures us that no plane contains all the points of
space?
55. Prove, using only the postulates, that every point is contained in at least
one plane,
56. Draw a single diagram which illustrates what all eight postulates assert.
Label the points in your diagram and then name one pair of skew lines,
57. What dees it mean to say that points A t B, and (7 are collinear?
5S, Does the ThreePoint Postulate eliminate the possibility of there being
tliree points in a plane that are collinear?
59. Assume that A, B, C, D, and E are five distinct points in the same plane
with no three of these points collinear.
(a) How many distinct lines are there each containing two of the five
points?
(b) How many lines determined by A, B, C t D, and E are there on each
point?
60. Let A, J3, C, D, and E he five distinct points in the same plane.
(a) If A, B, and C are on line 1, and C. D* and F are on line m, and l^m,
how many more lines are there each containing two of the five
points? How many lines are there altogether?
(b) How many distinct lines pass through point A? Point #? Point C?
6L If line I contains three distinct points A, B, and C and if points A and B
lie in plane a, what can you conclude?
62. Planes RST and S7T are distinct.
(a) Name a line that lies in both of these planes.
(b) Name two more planes determined by points H. S. 7", U.
6,1, tState what intersect means
(a) with regard to two lines,
(b) with regard to two planes,
64. Which of our Incidence Postulates "pushes" us off a plane into space?
Determine in Exercises 6571 if the given statement is true or false.
65. It is a theorem that every line contains at least two distinct points.
66. The Line Point Postulate assures us that there is just one line which
contains two distinct points.
67* It is a theorem that there arc at least throe distinct lines in every plane,
68. If a line intersects a plane not containing the line, the intersection is a
single point
69. It is a theorem that there are at least four distinct planes in space.
70. Our Incidence Postulates assure us of at least twelve distinct lines in
space.
71. The ThreePoint Postulate assures us that there are at least three dis
tinct points on every line.
SydCnrnberg/lXF.t.
Separation
and Related
Concepts
2.1 INTRODUCTION
In this chapter we list our second group of postulates. These postu
lates are concerned with points between other points, how a point sep
arates a line, how a line separates a plane, and how a plane separates
Space, The ideas these postulates convey arc very simple ones, and the
information they give can be easily determined by looking at pictures.
We list these postulates so that we can give a meaning to the rela
tionships of bctweenness and separation and to make it quite dear
what statements we agree to use in our proof & Later in the chapter we
use these relationships to make several definitions.
2.2 THE BETWEENNESS POSTULATES
In this section we state the Betwecnness Postulates. They are some
times called order postulates because they tell us how points are ar
ranged in order on a line. For example, if A* B, C are three distinct
points on aline, and if Bis between A and C, then we could say that the
points are arranged on the line in the Order A* then B, then C We could
also say that the order is C, then B t then A, (See Figure 21.)
a b C
• • 9
Frgrirc 21
42 Separation and Related Concepts
Chapter 2
In our formal geometry between and order, as they apply to points,
are undefined terms. We do not explain by a definition what it means
to say that point B is between points A and C Rather we accept the
following Betweenncss Postulates as formal statements of betweenness
properties.
The Betweenness Postulates
POSTULATE 9 (The ABC Betweenness Postulate) If point
B is between points A and C, then point B 1% also between C and A,
and all three points are distinct and collinear.
Note that the postulate does not allow us to say that "point B is
between points A and C" if the points are as shown in Figure 22a.
Ftpir* 2*1
Why? Similarly, in Figure 22b point B is not between points A and C.
Why? Figure 22c is a correct picture of what we mean when we say
"point B is between points A and C." We use the symbol ABC or
the symbol CBA when we wish to say that "point B is between points
A and C."
POSTULATE 10 (The ThreePoint Betweenncss Postulate) U
three distinct points are coUincar, then one and only one is between
the other two.
From this postulate it follows that if A, B t Care three distinct points
on a line, then BAQ or ABC, or BCA\ and if ABC, for exam
ple, then we cannot have BAC or BCA, From this postulate it also
2,2 The Betweenness Postulates 43
follows that betweenness for points in our format geometry is different
from betweenness on a circle. For example, if three people are seated
at a circular table, then most of us would agree that each of them is
between the other two. But we should also now agree, after accepting
Postulate 10, that betweenness around a table is different from be
tweenness in our formal geometry.
POSTULATE 11 {The LineBuilding Postulate) If A and B are
any two distinct points, then there is  point Xj such that X a is between
points A and B, a point Yg Sttdi that B is between A and Y lt and a point
Zi such that A is between Zj and B.
Figure 2*3 illustrates the meaning of Postulate 1 L We call tliis pos
tulate the LineBuilding Postulate because it enables us to prove that
there are infinitely many points on a line.
Zi a X x B Yi
4 * * • • * ^
Pigut* £3
From this postulate it follows that there is a point Yi on line AB
that is beyond B from A, a point Y 2 that is beyond Y x from B, a point
Y 3 that is beyond Yt from Y\, and so on (Figure 24).
Z\ A Xj B y, F 2 Yi Y±
4 • • • • • * — # — • » » « — ►
He..«v v. i
Tlius we are able to find on the line as many points as we choose,
3 or 30 or 3,000,000 or any natural numl>er you wish, that arc beyond
B from A. Similarly, we can find as many points as we choose that are
beyond A from B (Figure 25).
Z4 ?a Zi Z\ A Xi B Yt Y 2 Y 3 Y. t
4 — • • • • — • — • • • • • • • — • — • — • — •— . ■ ^
Figure 25
From this postulate it follows also that no matter how close
together two distinct points may be there is always a point between
them (Figure 26).
Z% Zi Zj A Jfj. X 2 Xs X* Xn B Yi Yi Y S
i •• • • • • — * — • • • • » * • • — • m » # 
Figure £6
44 Separation and Related Concepts Chapter 2
Does your intuition tell you that Postulate 11 builds every line com
pletely, that no "holes" could possibly exist in a line? Actually, Postu
late 11 assures 08 that there arc infinitely many points on every line and
furthermore, infinitely in any points between every two points of a line.
But it does not assure us that there arc no holes in a line. The postula
tional basis for lines is completed in Chapter 3, where we adopt a Ruler
Postulate. This postulate provides for the ideas of "continuity' 1 and
"infinite extent"
EXERCISES 2,2
In Exercises 122, indicate by 1" or F whether or not the statement is true
or false.
In Exercises 14, R, S, Tare pointi such that ftST.
1. /i„ $ ? T are collinear points.
2. R. 5, Tare distinct points.
3. TSR
4 SRT
In Exercises 5KJ, points A and B are on line m, point C is between A and
H and on line n. and m and n arc distinct Hues,
5. m and n intersect at A.
G. m and n intersect at B>
7. m and n intersect at C
8. A, B, C are collinear points.
9. ABC
10. CBA
hi Exercises 11—18, F, Q, R are distinct points and no one of them is between
the other two.
11. F, Q, R are collinear points.
12. P and Q are collinear points.
13. P and R are collinear points.
14 Q and R are collinear points.
15. There is one and only one plane that contains P, Q, and R,
2 J> The Betweenness Postulates
45
16. There is a point S such that QRS.
17. There Is a point 7 such that QTH,
IS. There is a point U such that UQR,
In Exercises 10223 R, S, Tare three distinct eollinear points, RST is false,
and 5R7" is false,
19. RTS
20. STR
2L TRS
22. TSR
Exercises 2332 refer to Figure 27 which shows three noncolhnear points
A, B, C and the three lines and the one plane that they determine.
Figure 2>7
23. Which Incidence Postulate asserts the existence of three noncoUinear
points?
24. Which Incidence Postulate assures us, once wo have points A, B t C,
dial lines AB t BC, CA exist?
25. Which Incidence Postulate assures us, once we have points A, B, G\
that there is a plane or containing them?
26. Which postulate assures us. once we have the points and the lines of
the figure, that there are points D t Z, F such that BUC % CEA,
AFH?
27. Explain why D and E are distinct points.
28. Why is DEA a false statement?
g& Why is EAF a false statement?
30, Which postulate assures us that diere are points G and H such that
EPC and DFIT?
31, Explain why // and C are distinct points.
32, Explain why AB and HG are distinct lines.
46 Separation and Related Concept* Chapter 2
■ Exercises 3335. The Incidence Postulates assert the existence of four non
coplanar points. Figure 26 suggests four noncoplanar points A, B, C„ D and
the four planes and sis lines determined by them.
Ffgun»4
33. Deduce from the Incidence and Betweenness Postulates that there is
at least one point in space that is not contained in any of the four planes
of the figure*
34. Deduce from tl»e postulates that there is at least one line in space that
is not contained in any of the four planes of the figure.
35. Deduce from the postulates that there is at least one plane in space that
is different from any of the four planes of the figure
Exercises 3642 refer to Figure 29 which shows a line and a portion of it
(the heavier part including the points C and D) called a segment (Segment
is defined in the next section,} Points X, Y, 2 as well as C and D are points
of this line as indicated in the figure.
< I £ 1 — I i ►
Figure 29
36. Are X and V points of the segment?
37. Is Z a point of the segmont?
38. Is CXD? 40. Is CZD?
39. Is CYD? 41. Is C C m
42. Try to write a definition of segment.
43. challenge pboblem. Let a line I be given. Deduce from the postu
lates that there are at least three different planes containing /.
44. challenge problem. Let P be u point in a plane a. Deduce from the
postulates that there are at least three dLUereuL lines in a passing
through P.
2.3 Using Batweenne&s to Make Definitions 47
2.3 USING BETWEENNESS TO MAKE DEFINITIONS
We now list several definitions that make use of the between ness
relation. Most of these are definitions of terms you have met in your
earlier work in geometry. However, it is the betweenness relation that
enables us now to make these definitions precise.
Figure 210 is a pic Lure of a set of points called a segment
A B
»■ ■ •
Figure 210
Definition 2,1 If A and B are any two distinct points, seg
ment AS is the set consisting of points A, B, and all points
between A and B. The points A and B are called endpoints
of AS
Sometimes a segment is called a line segment to avoid possible con
fusion with a "segment" of a circle. Be careful not to confuse the sym
bolAi? for a segment with the symbol AB for a line. The horizontal bar
in AB reminds us that a segment has endpoints, whereas the bar with
* — *
arrows in AB reminds us that a line has no endpoints.
We sometimes say that two distinct points determine a segment,
T hus, if A and B are two distinct points, then the segment determined
is AB. Similarly, if A,_B, C are three distinct points, then the segments
determined arc AB, BC, and CA.
1
#
Figure 2 II
Figure 21 1 is a picture of a set of points called a ray. As the figure
indicates, a ray has just the one endpoint A. Speaking informally we
might say that the ray continues without end from A through B in a
straight line. We express the idea of a ray formally in our next
definition.
Definition 2.2 If A and B are any two distinct points, ray
AM is the union of segment AB and all points X such that
ABX. The point A is called the endpoint of AB.
48 Separation and Related Concepts Chapter 2
In the symbol AB t the arrow reminds us that it is a ray. The arrow
in this symbol is always drawn from left to right regardless of the di
rection in which the ray points. However, do not confuse the symbol
AB with the symbol BA. Both represent rays, but they do not represent
the same ray, as seen in Figure 212, Observe that ray AB has endpoint
A, whereas ray BA has endpoint B,
A B
• •
VtayAB
R*y
A
tytiX
Figure 212
It is often convenient to speak about opposite rays. Figure 213
is a picture of what we mean by opposite rays. The figure suggests the
following; definition.
AC AB
B
Figure £13
Definition 2.3 If A is between B and C, then rays AB and
AC are called opposite ruys.
The betweenness relation CAB in this definition requires oppo
site rays to be collincar. Why? They mast also have a common end
point. Thus the two rays shown in Figure 214 are not opposite rays
even though they "point in opposite directions."
Q
—
Figure 214
2.3 Using Betweenness to Make Definitions 49
Note that the definition of opposite rays is equivalent to saying that
two distinct rays are opposite rays if they are colfinear and have the
same endpoint.
We often use the symbol "opp.AB" for the ray opposite AB. Thus,
in Figure 213, opp AB = AC since both symbols name the same set
of points, namely , the ray that is opposite AB.
The following summary should be helpful: If A and B are any two
distinct points on a line, they determine the six subsets of the line
shown in Figure 215, with subsets indicated by heavier marking. Be
sure to notice the differences among the symbols AB, AB, AB, and BA.
line AB = BA 4 £ fc
segment a_
B
ray AB 4 4 2
opp AB
myBA
A
oppBA 4— ♦
Figure i15
A segment has two endpoints and a ray has one endpoint. Some
times we wish to speak about those points of a segment or ray that are
not endpoints. Accordingly, we state the following definition .
Definition 2.4 The interior of a segment is the set of all
points of the segment except its endpoints. The interior of a
ray, also called a halftone, is the set of all points of the ray
except its endpoint.
50 Separation and Related Concepti Chapter 2
EXERCISES 2.3
L Use the ABC Betwecnticss Postulate to prove that every point of AB
is in AH.
2. Consider the following definition of a segment AB:
AB = (X : ,V £ AH. and X = A, or X = B, or AXB).
Is this definition equivalent to Definition 2.1?
3 Consider the following definition of a ray AB: If A and H are distinct
points, then ray AB consists of point A, point B> all points X such that
AXB, and all points V such that AB X Is this definition equivalent
to (Definition 2,2?
4. If A and B are distinct points, what is the intersection of rays AB and
BA?
5. If A and B arc distinct points, what is the union of AB and BA?
6. Write an alternate definition of segment in terms of intersecting rays.
7. Can two distinct rays have no print in common? Illustrate with a figure.
8. Can two distinct rays have exactly one point in common? Illustrate.
9. Can two distinct rays have infinitely many points in common? Illustrate.
10. Can the intersection of two distinct rays be a set containing two and
only two distinct points?
11. Can the intersection of two distinct rays be a ray?
12. Can two collinear rays have no point in common?
13. Is there a segment with no endpoint?
14. Is there a ray with no endpoint?
15. Is there a segment with no interior point?
16. Is there a ray with no interior point?
17> Is there a segment with exactly one interior point?
18. Is there a ray with exactly one interior point?
19. Is there a segment with exactly two interior points?
20. Is there a ray with exactly two interior points?
2L IF ABC, what is the intersection of AH and AC?
22. If ABC, what is the union of AB and AC?
2a (f ABC, what is the intersection of AB and BC?
24. If ABC, what is the union of "KB and B~C?
25. If ABC, what is the intersection of AB* and ^B?
26. If ABC, what is the union of AB and ~C$?
27. If ABC. if S is the Interior of BA t and if T is the interior of BC. what
is the intersection of S and 7"?
2.3 Using Betweenness to Mak# Definitions 51
28. If ABC, if S is the interior of BA, and if T is the interior of BC, de
scribe the union of S and T.
29. If 5 is the interior of a segment AB, what is the union of S and ABP
30. If S is the interior of a segment AB, what is the intersection of S and AS?
31. If A and B are distinct points of the segment HS, what is the union of
A3 and 55?
In Exercises 3241 , a subset of a Hne / is named. Points on line I are indicated
in Figure 216. Using tills figure, write a simpler name for the set. Exercise
32 has been worked as a sample.
t 2
5
C
fl
£ J?
Figure 21
32, CE H CE = CE
37. tf? n *S
.13, Ci U CE
38. ED U CF
34, Dp fl DA
39. imncF
35. o/^j DF H CF
36, ~CF U 5?
40. offi £? n
41. Mnopp
BE
CA
42. Prove that there is no segment on a line / which contains every point
of line /. (Hint: Let A~S be any segment on line I Show that there is a
point on line I that is not in AB,}
43. Iff and Q ore two distinct points on a line, does QP = opp PQ? Explain.
44. How many segments do three distinct points on a line determine? Four
distinct points? Five distinct points? (A given set of points determines
a segment if the endpoints of the segment are in the given set.)
45. Use your answers to Exercise 44 to predict how many segments six dis
tinct points on a lino determine. Test your prediction by counting.
46. (a} A, B f C t D are four distinct points on a line /. The conjunction of the
two sentences (1) and (2) is true if the appropriate symbol ( — , ^,
«*>) is supplied over each letter pair (BA). Copy these statements
and supply these symbols.
(1) BA contains points C and D, but BA contains neither of these
points.
(2) D belongs to BA, but C does not.
(b) Draw a sketch which shows the order of the four points A t B^QD
on I.
47. A % B, C. D are four distinct points on a line /. If C g DB, A £ DB,
C £ Bu. A £ BA draw a sketch which shows the order of the four
points on /..
52 Separation and Related Concepts
0,m:.:2
48, If FQ is opposite to PR* which one of the three points P, Q t R is between
the other two?
40. If opp RQ = RF, which one of the points P, Q, R is between the other
two?
2.4 THE CONCEPT OF AN ANGLE
The concept of an angle is a fundamental one in mathematics as
well as in the practical world of the house builder and the engineer.
There are several ways to think of an angle. One way is to think of an
angle as two noncollinear rays that have the same endpoint. A second
way is to think of an angle as the points on two noncollinear rays that
have the same endpoint. In one, an angle is a set of rays. In the second,
an angle is a set of points* In the following formal definition we agree
to think of an angle as a set of points.
Definition 2.5 An angle is the union of two noncollinear
rays with the same endpoint. Each of the two rays is called a
side of the angle. The common endpoint of the two rays is
called the vertex of the angle.
Notation. The angle formed by the rays AB and AC is denoted by
Z BAG. When using a symbol such as Z BAG it is important that the
middle letter denote the vertex of the angle. Since the union of A R and
AC is the same as the union of AC and AB, it should be clear that
Z BAC = Z CAB, (See Figure 217.) It should also be clear that many
different points may be used in identifying the same angle. Thus, if
AB = aS=AD and AE = A? = aS,
as indicated in Figure 218, then ZBAE = Z CAE = Z DAE =
Z RAF = Z BAG, and so on.
Figure 21"
Figure £18
If a figure or other information makes clear which rays are the sides
of the angle, we can denote Z BAE by simply writing Z A, However,
2.4 The Concept of in Angle 53
if there is more than one angle with vertex A as in Figure 219, we
would not know which angle is referred to by L A,
Figure M
It is often convenient to label an angle by indicating a numeral or
a lower case letter in its interior. Thus in Figure 219 we can write L 1
for Z DAC and Z r for Z BAD. When used, the arcs in the figure indi
cate the sides of the angle and the interior of the angle.
Sometimes we speak of the angle determined by two noncollinear
segments which have a common endpoint as indicated in Figure 2*20.
If A B and BC are the segments, then the unique angle determined by
them is ZABC
Figure %m
Observe in the definition of an angle that the sides of an angle are
noncollinear rays and therefore distinct rays. In some books this re
striction is not made, A special case of an angle results when iLs two
sides are "coincident." It is called the /em angle. Another special case
results if the two sides are distinct but coUinear. In this case the sides
of the angle are opposite rays and the angle is called a straight angle.
In Figure 221, Z ABC is a zero angle and ZDEF is a straight angle.
A
Figure Ml
54 Separation and Related Concept! Chapter 2
However, because zero angles and straight angles are not needed and
it is simpler not to consider them, they have been excluded from the
formal definitions in this book.
EXERCISES 2,4
1. Copy and complete the following definition: An angle is the [T] of two
[7] which have a common ondpoinl but do not lie on the same \t].
In Exercises 29, draw a picture to illustrate a set satisfying each of the given
descriptions.
2. Two distinct coplanar rays whose union is not an angle.
3. Two angles with the same vertex whose intersection is a ray.
4. Two angles with different vortices whose intersection is a ray.
5. Two angles whose intersection is a segment
6. Two angles whose intersection is a set consisting of cxacdy one point.
7. Two angles whose intersection is a set consisting of exactly two points.
8. Two angles whose intersection is a set consisting of exactly three points.
9. Two angles whose intersection is a set consisting of exactly four points.
10. If A, B t C are three distinct points on AB t if A, D„ E are three distinct
points on AD, and if A, B, D are noncollinear points, then the union ol
18 and A D is an angle which we may indicate as L BAD. Using A, B,
C, D, E, write the other possible symbols for this angle.
11. In the following figure , the capital letters denote points and the lower
case letters demote angles. Using three capital letters, write another
name for each angle denoted by a lower case letter.
(a)*=ZQ (d)«=Z[?3
(b) y= im (e) r,= Z[?]
(c) %= ZQ (f) w= L\T\
2.4 Th* Concept of an Angia 55
12. How many angles are determined by three distinct coplanar rays having
a common endpoint if no two of the rays are opposite rays? By four dis
tinct rays? By five distinct rays? (A given set of rays determines an angle
if the sides of the angle are elements of the given set.)
13. Would your answers to Exercise 12 be different if the rays were not all
in the same plane?
14. In the figure below, AB and AC are opposite rays. How many angles do
the four rays determine? Name them.
15. If two distinct lines intersect, how many angles are determined? (A
given set of lines determines an angle if each side of the angle is con
tained in one of the lines of the given set.)
16. If three distinct coplanar lines; intersect in a common point, how many
angles do they determine?
17. Name all the angles determined by the segments shown in the figure.
How many are there? Which angles can Ijc named by using only the
vertex letter?
18. Using letters in the figure in Exercise 17, name the
(a) angle with vertex P in four different ways.
(b) ray that is the intersection of APRS and I QRS.
(c) ray that is the intersection of APSR and / QSR.
(d) segment that is contained in the intersection of Z PQR and Z SRQ,
(e) point that is not in FR but is in the intersection of I.RPQ and
APRS.
(f) three distinct rays contained in the union of Z PRS and Z QRS.
56 Separation and Related Concept*
Chapter 2
Exercises 1930 are informal geometry exercises. The formal development
of angle measure appears in later sections. Using Figure 222, which shows
degrees on a protractor, compute the number named in the exercise. Exer
cise 19 has been worked as a sample.
figure 2*2a
19. mlDAF = 55
20. ml BAD
21 mlDAH
22. mlHAE
23. mil AH
24. mllAE
25. mlEAD
2«. ml IAD
27. mllAE + tnlEAD
28. mlDAH  mlllAE
29. ml DAE
30. ml DAE + mlDAF
31. Eveiy triangle is the union of three segments, but not every union of
throe segments is a triangle. Write a definition of a triangle that you
think will "hold water." (The formal definition of triangle appeals in a
later section.)
2.5 THE SEPARATION POSTULATES
Tn this section we state three more postulates involving die idea of
betweenness, which we call the Separation Postulates. These postulates
tell us how a point separates a line, how r a line separates a plane, and
how a plane separates space. Although the ideas are very simple, we
cannot prove them from the postulates thus far agreed on. In order to
facilitate their phrasing, wc introduce the idea of a convex set.
Definition 2.6 A set of points is called convex, if for every
two points F and Q in the set, the entire segment F^Jis in the
set. The null set and every set that contains only one point
are also called convex sets.
2.5 The Separation Postulates 57
The definition implies that if P and Q are any two points of a con
vex set T, then PQ C T. A line, a ray* and a segment are examples of
convex sets of points as Figure 223 suggests. Is a plane a convex set of
points?
= / +
+1
_ — *A
FQt=ABm
PQ^Alti
Q
Figure £23
The interior of a triangle is a convex set Figure 224 shows two
choices. Pi, P* and Q r , Q 2y for points P and 9, respectively, in the in
terior of the triangle and in each case the entire segment PQ is con
tained in the set Is the interior of the circle shown in Figure 224 a
convex set? (We define interior of triangle, circle, and interior of circle
formally in later sections,)
Fisure 224
However, none of the sets shown in Figure 225 is a convex set In
each instance it is possible to find points Pand Q such that not all of the
segment PQ is contained in the set.
J
1
p
/
Figure 225
58 Separation and Related Concepts Chapter 2
Let S be a convex set, T a convex set, and R the intersection of the
two sets. If R is the null set or if ft consists of a single point, why is R a
convex set? Suppose, then, that R consists of more than one point. Let
P and Q be any two distinct points in H. Draw an appropriate figure
to suggest the sets S T % R and the points P and Q, Then answer the fol
lowing questions.
1. Why is P in the set S? In the set T?
2. Why is pin the set S? In the set T?
3. Why is P£ in the set S? In the set T?
4. Why is PQ in the set R?
5. Does this prove that R is a convex set?
6. Does this prove the following theorem?
THEOREM 2.1 The intersection of two convex sets is a convex
Figure 226 suggests the first of our Separation Postulates. It shows
point A dividing or separating line / into two convex sets T S and T,
4 * • • • ► t
B A C D
Figure 336
No point of / is in both S and T, Point A is the only point on line I that
is in neither S nor T, If points B and C are in different sets (that la, if
B £ § and C £ T, or if C £ S and R £ T) t then EC contains point A*
On the other hand, if points C and D are in the same set (that is, if
C £S and D £ S, or C £ T and D £ T ), then CD does not contain
point A.
We state these ideas formally in our next postulate.
POSTULATE 12 (The Line Separation Postulate) Each point
A on a line separates the line. The points of the line other than the
point A form two distinct sets such that
1. each of the two sets is convex;
2. if two points are in the same set.
then A is not between them;
3. if two points are in different sets,
then A is between them.
2.5 The S*pa ration Postulates 59
Definition 2.7 Let a line I and a point A on I be given,
1. The two convex sets described in Postdate 12 are called
hairline* or sides of point A on line I; A is the endpoint of
each of them.
2. If C and D arc two points in one of these sets, we say that
C and D are on the same side of A, or that C is on the
Dside of A, or that D is on the Cside of A.
3. If B is a point in one of these sets and C is a point in the
other set, we say that B and C are on opposite sides of A
on line / or that B and G are in the opposite halftones of l
determined by the point A.
Note that a halfline is a ray with the endpoint omitted. Although it
is convenient to think of a halfline as Juicing an endpoint, remember
that a halfline does not contain its endpoint. (This use of "have" should
not bother yon since you have friends, but you do not contain them.)
Note also that opposite hairlines are collineur and that they have the
same endpoint.
Since there are infinitely many lines in space that contain any given
point, it follows that although a point has only two sides on any given
line, it has infinitely many sides in a plane or in space. Normally, we
do not speak of the sides of a point except when a given line contains
the point and we are discussing the sides of the point on that line.
Postulate 13 describes how a line separates any plane containing
die line. In Figure 227, we see that the points of plane « that do not
lie on line / are separated into two sets:
(1) those points that are on the Bside of line h
(2) those points that are on the Csidc of line /.
«
Figure 127
60 Separation and Related Concepts Chapter 2
No point is in both of these sets and no point of line I is in either of
the two sets. If points B and Care in different sets, then 2?C intersects I.
On the other hand, if points C and D are in one set, then CD does not
intersect line L We state these ideas formally in our next postulate.
POSTULATE 13 ( The Plane Separation Postulate) Each line I
in a plane separates the plane. The points of the plane other than the
points online / form two distinct sets such that
1. each of the two sets is convex;
2. if two points are in the same set, then
no point of line I is between them;
3. if two points are in different sets, then
there is a point of line I between diem.
Definition 2.8 Let a plane a and a line / in a be given.
1. The two convex sets described in Postulate 13 are called
halfplanes or sides of / in plane a; l is the edge of each of
them.
2. If C and D are two points in one of these sets, then we say
that C and D arc on the *ame side of / in plane a t or that C
is on the D side of t t or that D is on the Cstde of I or that C
and D are in the same h airplane.
3. If B is a point in one of these sets and C is a point in the
other set, wc say that B and C are on opposite sides of / in
plane a or that B and C are in the opposite halfplanes of a
determined by the line L
Note that a halfplane does not contain its edge and that opposite
halfplanes are coplanar and have the same edge. Furthermore, since
there are infinitely many planes in space that contain any given line, it
follows thai although a line has only two sides in any given plane, it has
infinitely many sides in space. Normally, we do not speak of the sides
of a line except when some given plane contains the line and wc arc
discussing the sides of the line in that plane.
Figure 228 suggests the last of the Separation Postulates. The
figure shows how a plane a separates the points of space that do not lie
in plane a into two sets:
(1) those points that arc on the Bside of plane a;
(2) those points that are on the Cside of plane a.
No point is in both of these sets and no point of plane a is in either
of the two sets. If points B and C are in different sets, then BC inter
2,5 Th* Separation Postulates 61
Figure £28
sects plane a. If points C and D are in the same set, then CD does not
intersect plane a.
Although these space separation ideas can be proved using the
postulates introduced up to this point we shall summarize and state
these ideas formally as a postulate in order to simplify and shorten our
formal development of geometry. Note the similarity of the Space
Separation Postulate to the Plane Separation Postulate.
POSTUIATE 14 (Tke Space Separation Postulate) Each plane
tit in space separates space, The points in space other than the points in
plane a form two distinct sets such that
1. each of the two sets is convex;
2. if two points arc in the same set, then
no point of plane a is between them;
3. if two points are in different sets, then
there is a point of plane a between them.
Definition 2.9 Let a plane a tie given.
1. The two convex sets described in Postulate 14 are called
half spaces or aides of plane a and plane a is called the face
of each of them.
2. If C and D are any two points in one of these sets, then we
say that C and D arc on the same side of a., or that C is on
the Dside of a, or that D is on the Cside of a, or that C
and jD are in the same halfspaec,
3. If B is a point in one of these sets and C is a point in the
other set, then we say that B and C are on opposite sides
of a or that B and C are in opposite halfspaces.
62 Separation and Refated Concepts
Chapter 2
Note that a halfspace does not contain its face and that opposite
halfspaces have the same face. Furthermore, although a point or a line
has infinitely many sides in space, a given plane has only two sides in
space.
The statements of the Separation Postulates and the definitions
accompanying them are lengthy. However, the ideas they convey arc
simple and can be easily described by means of figures. Briefly, the
Separation Postulates tell us that a point separates a line into two half
lines; that a line separates a plane into two halfplanes; that a plane
separates space into two halfspaces; and that these sets are convex.
Notation. Figure 229 shows point A on line I and two halftones de
termined by point A, Wc may denote the halHine on the Cside of A in
c ►
line / by the symbol U or by the symbol AC. Similarly, the halfline on
the Bside of A may be denoted by I2 or by AB. Be careful to note the
difference between the symbol AB for ray AB and the symbol AB for
hairline AB. Does AB = a3? Explain.
p
C
■*/
Figur«S 28
Figure 230 shows line I in plane a and the two halfplanes which
line I determines. We may denote the halfplane on the /Jside of line 1
in plane a by ffi, and the halfplane on the Cside of line I in plane a by
a 2 . Similarly, if S represents the set of all points in space, we may de
note the two halfspaces into which space is separated by a plane with
the symbols Si and S2.
D«
Figure 230
2.5 The Separation Postulate* 63
We make use of the Separation Postulates to prove the next two
theorems,
THEOREM 2,2 If m segment has only one endpoint on a given
line, then the entire segment, except for that endpoint, lies in one
half plane whose edge is the given line,
RESTATEMENT:
Given: Line I and segment AB in plane a such that / and AB have
only the point A in common,
To Prove: If X is any point of AB such that AXB, then X is on
the Bsade of Mn plane a.
Figure 231
Proof.
1, X is a point such that AXB.
2, A, X, B are distinct points,
3, I does not intersect XK
4, X and B are not on opposite sides of I
5, X and B are on the same side of /,
1. Given
2. Why?
3. Why?
4. Why?
5. Plane Separation
Postulate
We have shown that X, which is any point of AB except point A or
point B, is on the Bside of t in plane a; hence every point of AB, except
point A, is on the Bside of / in plane a.
THEORFM 2.3 If the intersection of a line and a ray is the end
point of the ray. then the interior of the ray is contained in one half
plane whose edge is the given line,
RESTATEMENT:
Given: AB and line / i ntersect in j ust the point A in plane a.
To Prove: All points of AB are on the Bside of / in plane a.
64 Separation and Related Concepts Chapter 2
Figure 238
Proof: Let X be any point of AB such that ABX.
1. All the points of All, except A, lie on the 2Mde of / in plane
cl Why?
2, X £ I Why?
& I does not intersect BX. Why?
4, X is not on the opposite side of I from B, Why?
5, X is on the same side of t as B, Why?
Since X is any point such that ABX, it follows that crop BA Ues
entirely on the Bside of I Since AB is the union of opp BA and ?0f with
A deleted, it follows that A ft is on the Bside of I
EXERCISES 2.5
1. Prove that the intersection of any three convex sets is a convex set,
2. There is a theorem which "extends" Theorem 2.1 to any numl>er of sets.
Stale this theorem.
3. Is a ray a convex set?
4. Is the interior of a ray ft convex set?
5. Consider the following definition, of a ray: Ray AS consists of point A
and all the points on the Bside of A on line AB. Define opp ~AB in a
similar manner.
6* Complete the following: Let SI be a segment. Then AB — AB f\ BA.
Since AB and BA are convex sets it follows from [7] that [T] is a convex
set.
7. Let I be any line and let P and Q be any two distinct points on /. If X
is any point between P and (>, is X on line I? Why? Does this prove that
PQ is contained in I? Does this prove that every line Is a convex set?
8* Let a be any plane and let Pund Q be any two distinct points in plane «.
(a) Is PQ in plane a? Why?
(b) Is PQ* subset of ?$* Why?
2.5 The Sepa ration Postulates 65
(c) Is PQ a subset of plane «? Why?
(d) Does this prove that every plane is a convex set?
9. Hie interior of each circle shown in the figure below is a convex set
(a) Is the intersection of these interiors a convex set? Why?
(h) Is l he union of these interiors a convex set?
10, Draw two circles in such a way that the union of their interiors is a
convex set,
*— • * — »
Exercises 1 115. In Figure 233, lines AB and CD intersect at E so that
AEB and CED. The Dside of a2 has been shaded.
Figure £33
11. Copy the figure and shade the Bside of
12. Why are the two shaded half planes coplnnar?
13. Why is the intersection of these two half planes a convex set?
14. Describe in your own words the intersection of the two shaded half
planes.
15. Does your description in Exercise 14 suggest a definition of "interior of
an angle?"
16. Let P, Q, and R be distinct points on a line L with ft and Q in the same
hatfline with endpotnl P, On the basis of the given information is it pos
sible that P is between Q and R? That Q is between H and P? That R is
between P and Q?
17. Let F, Q, R be distinct points on a line /, with R and Q in opposite half
lines with endpoiut P. On the basis of the given information, is it pos
siWe that JP is between Q and R? That Q is between R and F? That R is
between P and OP
66 Separation and Related Concept! Chapter 2
18. Is the union of two opposite halflmes a line? Explain.
19. Is the union of two opposite halflines a convex set?
20. Is the union of two opposite rays a line? Is their union a convex set?
21. In what respect does the set of points in ray AB differ from the set of
o— ►
points in halflme AB?
22. Is the interior of ray AB the same set of points as halfline a2?
23. Is the u niou of two opposite h airplanes a plane? Ls their union a convex
set?
24. Line / lies in plane a, Point A is on one side of / in plane n. Point B is in
plane a and is not on the A side of I and is not on the opposite side of /
from A. Make a deduction,
25. A, B, C, X arc lour distinct points on line m, B and A are on opposite
sides of A* and C is on the A side of X. Draw a conclusion about points B
and C.
26. E, F y C are three distinct points in plane a. E and F are on opposite sides
of line n in plane a. If E and C are on opposite sides of line ft, what con
clusion can you draw with regard to points G and F?
27. From which postulate may we infer that a given line in a given plane
Ixas only two sides?
28. Explain why the following statement is true: If J* and Q are any two dis
tinct points in halfplane on, then FQ is in a*.
29. A halfplane is an example of a "connected region," A line in a plane
separates the points of the plane not on this line into two connected
regions. Into how many distinct connected regions do two distinct inter*
seeting lines separate the remaining points of the plane that contains
them?
30. Into bow many distinct connected regions do three distinct coplanar
lines separate the remaining points of the plane that contains them if no
point lies on all three lines and if each two of the lines intersect?
Zh Into how many distinct connected regions do four distinct coplanar
lines separate the remaining points of the plane that contains them if no
three of the lines contain the same points and if each two of Lhe lines
intersect?
32. Use your answers to Exercises 2931 to predict the number of distinct
connected regions into which five distinct coplanar lines separate the
remaining points of the piano that contains them if no three of the lines
contain the same point and if each two of the lines intersect
33. Can three distinct coplanar lines be situated so as to separate the re
maining points of the plane that contains them into three distinct con
nected regions? Four distinct connected regions? Five distinct con
nected regions? Six distinct connected regions? Seven distinct
connected regions? More than seven distinct connected regions?
2,6 Interiors and Exteriors of Angles €7
Figure 2.M
34. Draw a figu re for each part of Exercise 33 to which yon answered "Yes.
35. Figure 234 shows two distinct
planes ft and /? intersecting in
a line t The line Z is the edge of
taw many different halfplanes
represented in the figure?
36. Name two distinct halfplanes
represented in Figure 234 that
are coplanar.
37. Name two distinct halfplanes
represented in Figure 234 that
are not coplanar.
38. How many different pairs of
halfplanes in Figure 234 are
not coplanar?
39. Chic plane separates the rest of space into two connected regions. Into
bow many distinct connected rejpons do two distinct intersecting planes
separate the rest of space?
40. Into how many distinct connected regions do three distinct intersecting
planes separate the rest of space if no line lies in all three of the planes,
if every two of the planes intersect, and if each plane intersects the line
of intersection of the other two pla
41. L'se your answers to Exercises 39 and 40 to predict the number of dis
tinct connected regions formed by four distinct intersecting planes if
no three of these planes contain the same line, if each two of these
planes intersect, and if each plane intersects each line of intersection
formed by two of the other planes.
42. challenge problem. Construct a model to represent the situation of
Exercise 41. Count the number of distinct connected regions formed.
How does this number compare with your prediction?
43. challenge problem. Extend the result of Exercise 41 to five planes.
2.6 INTERIORS AND EXTERIORS OF ANGLES
In Section 2.5 we introduced the concept of separation. A line in a
plane separates the points of the plane not on the line into two half
planes. A pie rure of an angle suggests that an angle separates its plane.
Indeed, if plane a contains tLABC, then all the points of at that are not
points of L A BC make up two sets, one called the interior and the other
the exterior of Z ABC. We shall state carefully what we mean by these
terms.
Separation and Related Concepts
Chapter 2
One of the simplest ways ly think of the interior of an angle is as the
intersection of two halfplanes associated with the angle, For /.ABC
%ur*235
these haifplanes are the Csitle of AS and the Aside of BC, as indicated
in Figure 235. Our formal definition is as follows:
Definition 2.10 The interior of an angle, sa y £ABC t is
the intersection of two haifplanes, the Cside of AB and the
Aside of BC.
Figure £38
Definition 2.11 The exterior of an angle is the set of all
points in the plane of the angle except those points on the
sides of the angle and in its interior.
Figure 236 illustrates both Definitions 2. 10 and 2.1 1.
Tile following theorem is easy to prove using Definition 2.10 and
Theorem 2.3,
2.6 Interiors and Exteriors of Angles 69
THEOREM 2.4 If P is any point in the interior of Z ABC, then
the interior points of ray BP are points of the interior of Z ABC
RESTATEMENT:
Given: Z ABC with P a point in the interior of Z ABC.
To Prove; B?is in the interior of /.ABC
Proof: By definition of the interior of L ABC,
P is on the Cside of AJB and on the Aside of
BC. By Theorem 2.3, BP is on the Cside of AB
* — » o •
and on the A side of BC. Therefore BP is in
the intersection of these two halfplanes which,
by definition, is the interior of /ABC.
c
Figure 237
EXKRCISE5 2.6
L Copy and complete the following definition.
The interior of /PQR is the [7j of the haffplane lhatis the
Pside of [f] and the halfplane lhat is the \?j oi PQ.
Exercises 27 refer to Figure 238.
2* Which of the labeled points are in the interior of Z ABC?
& Which of the labeled points arc in the exterior of Z ABC?
4. Which of the labeled points are not in the interior of L ABC?
5. Which of the labeled points are in the interior of Z GBR?
6. Which of the labeled points are in the interior of Z G7J/JP
7. Which of the labeled points are in the interior of Z CBG?
70 Separation and Related Concepts Chapttr 2
§, Is the vertex of an angle u point of the interior of the angle? Explain.
9. Is the vertex of an angle a point of the exterior of the angle? Explain,
10. Is B a point of the interior of Z ABC? Explain.
11. Is B a point of the exterior of LABC? Explain.
12. Suppose that A, B, C, D are four noncoplanar points. Is it possible that
D k a point of the interior of <LABC? Explain.
13. Suppose that A, B t C, D are four noncoplanar points. Is it possible that
D is a point of the exterior of £ABC? Espial
■ Exercises 1420 refer to Figure 2*39.
Kipire 23S
14. Make a copy of Figure 239 and shade the A^side of F.B with vertical
raw ff and the £sfde of AB with horizontal rays ^,
15. Which angle has only vertical shading in its interior?
16. Which angle has only horizontal shading in its interior?
17. Which angle has both kinds of shading in its interior?
IS, Which angle has no shading in its interior?
19. Is the intersection of the I wo shaded portions the interior of any of the
angles shown in the figure? If so, name the angle(s).
20. Is the union of the two shaded portions the exterior of any of the angles
shown in the figure? The interior?
21. Does au angle separate the points of its plane not on the angle into two
connected regions that do not intersect?
22. What axe the connected regions of Exercise 21 called?
23. Draw a picture of a n angle. Mark two points Pand Q hi the exterior of
the angle. Does PQ intersect the angle?
24. Does your answer in Exercise 23 depend on your choice of P and Q or
would the answer he t he same for every choice of P and Q?
25. Is the exterior of an angle a convex set?
26. Draw a picture of an angle. Mark two points P and Q such thai P is
in the interior of the angle and Q is in the exterior. Does PQ intersect
the angle?
2,6 Interiors met Exteriors of Angles 71
27. Does your answer in Exercise 26 depend on your choice of F and Q t or
would the answer be the same for every choice of P and Q if P is chosen
in the interior of the angle and Q in the exterior?
28. challenge PimiiLEM. It can he proved that if P is a point in the in
terior of an angle and Q is a point in the exterior,, then PQ intersects the
angle. The proof is difficult and there are essentially five cases to con
sider as indicated in Figure 240 where £ ABC is the angle, P is a point
in the interior of the angle, and Qi, Qs, Q& Q± t Q& are points in the ex
terior of the angle such that
■i — * *— i ■»
(1) Qi is any point on the A side of BC and on the opposite side of AS
from C.
(2) Q 2 is any point on Vpp B(
^ y
(3) Q3 is any point on the opposite side of AB from C and on the oppo
site side of BC from A*
{4} Qa is any point on &pp BA,
(5) Qs, is any point on the Cside of AB and on the opposite side of BC
from A.
*
/
■%
Wjjure 240
Following is an outline of a proof for case 1, a proof that if Qi is any
point on the Axidc of BC and on the opposite side of AB from C, then
~F^fi intersects LABC. Answer the "Whys" in this outline,
i. P and C are on the same side of A H. Why?
2. C and Qi are on opposite sides of AB. Why?
3. P and Qi are on opposite sides of AB* Why?
i — »
4. There is a point R between P and Qi on line BA, Why?
<■ — *
This means that PQi intersects AB in an interior point of PQi . We wish
now to prove that PQ\ intersects BA rather than Qpp BA.
72 Separation and Ralated Concepts Chapter 2
5. P Is oil the Aside of M Why?
6. Q t is on the Aside of K?. Why?
7. The Aside of BC is a convex set Why?
B. AH points of FQx are on the Aside of KJ. Why?
Therefore the point fl is on the Aside of BC and on M. Therefore R
is on BA, and hence H is a point of I ABC. Therefore PQi intersects
labc.
29. challenge pboblkm . Let the situation of Exercise 28 be given. Prove
that PQ 2 intersectsTJA.
30. CHALLENGE problem. Let the situation of Exercise 28 be given. Prove
that P{) s intersects L ABC.
31. Draw a picture of a n angle. Mark two points F and Q in the interior of
the angle. Does PQ intersect the angle?
32. Docs your answer in Exercise 31 depend on your particular choice of
P and Q?
33. Is the interior of an angle a convex set?
34. challenge pkoblem. Use Theorem 2.1, the Plane Separation Postu
late, and the definition of interior of an angle to prove that the interior
of an angle is a convex set.
35. On the basis of your experiences in informal geometry try to write a
definition of a triangle, thinking of it as a set of points. (Remember that
a segment is a set of points and that the union of several sets of points
is a set of points.)
36. On the basis of your experiences in informal geometry try to write a
definition of a quadrilateral.
2.7 TRIANGLES AND QUADRILATERALS
Next to segments and angles perhaps the simplest geometrical fig
ures are the polygons, and the simplest polygons are the triangles. You
all know what a triangle looks like. It has three sides and three angles,
A drawing of a triangle (Figure 211) shows its three sides, which are
the segments AB, BC, and GA.
2.7 Trianghw and Quadrilaterals 73
On the other hand, the angles of a triangle are not shown completely
in a picture of the triangle. In Figure 242, however, there are pictures
showing the tingles of a triangle. A picture that does show all the angles
of a triangle is not a picture of what is usually meant by a triangle.
its uigto*
Figure 242
The following definition is worded carefully so that it says exactly
what we want it to sav.
Definition 2.12 If A, B f C are three noncollinear points,
then the union of the segments A B, EC, CA is a triangle.
Notation. The triangle which is the union of the segments AB, BC,
CA is denoted by AABC
x\n alternate definition which you might prefer is as follows.
► A triangle fe the union of the three segments determined by
three noncollinear points.
This is an acceptable definition since we have agreed (in Section
2.3) that "if A, B. C are three distinct points, then the segments de
termined are AB, BC, and CA."
74 Separation and Related Concepts
Chapter 2
Definition 2J3 Let A AiJC be given.
1. Each of the points A, B. C is a vertex of A ABC
2. Each of the segments AB, W, CA is aside of A ABC.
3. Each of the angles L ABQ Z BCA, L CAB is an an^e of
AABC.
4. A side and a vertex not on that side are opposite to each
other.
5. A side and an angle arc opposite to each other if that side
and the vertex of that angle are opposite to each other.
Note that a triangle contains its vertices and its sides but that it does
not contain its angles. An angle of a triangle is not a subset of the tri
angle. It is customary to speak of the "angles of a triangle" or "the
angles determined by a triangle," but it is incorrect to refer to them as
"the angles contained in the triangle." Remember that whereas a tri
angle has angles, it does not contain them. (In this connection it might
be helpful to think of a farmer who has farms and barns but does not
contain them, or of a man who has a car and a house and lot but does
not contain them!)
Figure 243 illustrates the interior of a triangle and the interiors of
the angles of a triangle*
AA HC, l~A t interior of L A
AABC, jLB, interior of Z_B
AABC, LC, interior of £C
Figure £43
AABC, interior AABC
2.7 Triangle* and Quadrilateral* 75
Definition 2.14 The intersection of the interiors of the
three angles of a triangle is the interior of the triangle.
Definition 2.15 The exterior of a triangle is the set of all
points in the plane of the triangle that are neither points of
the triangle nor points of the interior of the triangle.
In Figure 244 the interiors of Z A and L B are shaded. Note that
the intersection of these interiors is the interior of the triangle.
figure 2M
This Suggests the following theorem,
THEOREM 2. 5 The i ntersecrion of I he interiors of t wo angles of
u triangle is the interior of the triangle.
Proof: Let A ABC be given. I^et X be any point in the intersection of
the interiors of any two of its angles, say L A and Z B.
X is on the Bside of AC because it is in
the interior of LA.
X is on the A side of BC because it is in
the interior of Z B.
X is in the interior of Z C because it is on
die Bside of AC and the Aside of BC.
This proves that if a point is in the interiors of two angles of a tri
angle, then it is also in the interior of the third angle. Therefore the in
tersection of the interiors of two angles of a triangle is contained in the
76 Separation and Related Concepts Chapter 2
interior of the triangle, Since the interior of the triangle is the inter
section of the interiors of all three angles, it follows that the interior of
the triangle is contained in the intersection of the interiors of any two
of its angles. Therefore the interior of the triangle is contained in, and
also contains, the intersection of the interiors of any two of its angles.
Therefore the interior of the triangle is the intersection of the interiors
of any two of its angles.
Consider any three n on colli near points A, B, C in a plane a and a
line I in a which does not contain A or B or C but which does contain
an interior point of AC as shown in Figure 24*5.
Figure fc45
L B is either on the Cside of I or on the Aside of / in plane a.
Why?
2. If B is on the Csidc of I, then I intersects AB in a point in the
interior of AB, Why? Does I intersect BC in this case? Why?
3. If B is on the Aside of /, then I intersects BT in a point in the
interior of BC. Why? Does / intersect AB in this case? Why?
4. Does this prove the following theorem?
THEOREM 2,6 If a line and a triangle are coplanar, if the hue
does not contain a vertex of the triangle, and if the line intersects
one side of the triangle, then it also intersects just one of the other
two sides.
Triangles have three sides. Quadrilaterals have four sides. By quad
rilaterals we mean some., but not all, plane figures that are made up of
four segments as suggested by Figure 246.
Try writing a definition of quadrilateral before reading further.
Then compare your definition with the following definition which we
adopt for our formal geometry.
2.7 Triangles and Quadrilaterals 77
Quadrilateral*
Not quadrilftt*rjiln
Figure 246
Definition 2. 16 Let A, B T C, D be four coplanar points such
that no three of them are collinear and such that none of the
segments AB, ffll, CD, DA intersects any other at a point
which is not one of its endpoints. Then the union of the four
segments AB, B£, UD, DA is a quadrilateral. Each of the four
se^nents is a side of the quadrilateral and each of the points
A, B, C, D is a vertex of the quadrilateral.
Is it possible that four given points are the vertices of more than
one quadrilateral? Figure 247 shows that this is indeed possible. Think
of the figure as four different pictures of the same four points. The sec
ond, third, and fourth pictures show different quadrilaterals with the
same vertices.
p
A
E
ASDC
a urn
ADBC
Figure 247
To name a quadrilateral using the names of its vertices, and to do
it so that we know which segments are its sides, the names of the ver
tices are so arranged that { 1) letters adjacent to each other in the name
78 Separation and Related Concepts Chapter 2
of the quadrilateral are names of the endpoints of a side of the quadri
lateral and (2) the first and last letters in the name of the quadrilateral
arc names of the endpoints of a side of the quadrilateral.
Note that if AB CD is a quadrilateral, then BCDA is the same quad
rilateral. Give several additional names for this quadrilateral. Notice
also that ABCD and ACBD are not names for the same quadrilateral.
EXERCISES 2.7
1. Copy and complete the definition: If A, B, C are three [?] points, then
is the triangle denoted by A ABC.
2. Name the side which is opposite lo /. ABC in A ABC.
% Name the angle which is opposite to side AB in A AUG'.
In Exercises 410, let A ABC be given. State whether or not the given set
is a subset of A ABC,
4. AABC 8. AB U BC J CA
5. Interior af / ABC 9. AB f BC ft 25
6. Interior of AABC 10. {A, B s C}
7. AB
In Exercises 1115, let AABC be given. Let S denote the union of AABC
and its interior. State whether or not the given set is a subset of S.
11. Interior of £ABC
12. (Interior_of_/ ABQ D (Interior of IBCA)
13. [AB.BC, CA)
14. AB HBCnCA
15. AB n (Interior of A ABC)
In Exercises 1619, let AABC be given. Let II U H 2x H 3 denote the follow
ing halfplancs, respectively: the A side of BC, the Bside of CA, and the
Cside of AB. Express the given set in terms of these halfplanes.
16. Interior of £ABC 18. Interior of ZCAB
17. Interior of LBCA 19, Interior of AABC
In Exercises 2024, let the same situation as in Exercises 1619 be given.
State whether or not the given set is a subset of the exterior of AABC.
20. Tlie interior of the ray 22. BC
opposite to BC 23, BC
21. The ray opposite to BC 24. The opposite halfplane front //j
2.7 Triangles and Quadrilaterals 79
25. Draw a picture of a triangle and a line so lhat
(a) their intersection is one point;
(b) their intersection is exactly two points;
(e) their intersection consists of more than two points.
26. Gun a line intersect a triangle in exactly one point which is not a vertex?
Illustrate.
27. What is your answer to lixercise 26 if the line and triangle arc contained
in the same plane?
28. Can a line intersect a triangle in exactly three points?
29. Name all of the triangles shown in the figure.
D
30. If A , B t C are the vertices of a triangle, prove that there is a point of L A
that is not a point of the triangle.
31. If A, H f C are noncollincar points, is the following statement true?
AABC = {I A n IB) U {I A n Zqj(ZBn ZC>
32. Let A ABC with points P and {> such that AFB and A QC as in the
figure be given. To prove that A ABC is not a convex set of points, it is
sufficient to show that there is a point X of fQ which is not contained
in A ABC. Let X be a point of TQ such that PJEQ,
A
(a) X £ AC.
(b) X € AB.
(c) X £ BC. Why? {Hint: Use Tlicorem 2.6.)
Does this prove that B triangle is not B convex set?
33. Is the exterior of a triangle a convex set? Illustrate with a figure.
34, Is the interior of a hi angle a convex set? Kxplain why.
80 Separation and Related Concepts Chapter 2
■ For Exercises 3538 draw a triangle, A ABC.
35. Mark a point that is in the interior of Z A and in the exterior of the
triangle.
30* Mark a point that is in the interior of Z A but is neither In the interior
nor the exterior of the triangle.
37. Mark a point in the exterior of the triangle that is not in the interior of
any of the angles of the triangle.
38, Can you mark a point that is in the interior of Z A and is in the interior
of ZB but is not in the interior of ZC? Why?
■ In Exercises 3942, let quadrilateral ABCD be given.
39, Is it possible that D is an element of BC?
40, Is it possible that D is an element of the interior of L BAC?
41. Is it possible that D is an element of the exterior of Z BAC?
42. Is it possible that the interior of A ABC is a subset of the interior of
A A DC?
43. Prove that every plane contains a quadrilateral,
44, Prove that any three vertices of a quadrilateral are the vertices of a
triangle.
2.8 PROPERTIES OF EQUALITY AND NUMBER OPERATIONS
In preparation f or Chapter 3, which involves some algebra, we con
chide Chapter 2 with several sections devoted to a review of elemen
tary algebra. Chapter 3 contains many equations, that is, statements
of equality. Following arc some basic properties of equality and num
ber operations that are useful in working with equations involving
numbers.
Substitution Property of Equality. In any statement about some
thing (physical object, number, point, line, idea, etc) one of the
names for that thing may be replaced by another name for the same
thing. If the original statement is true, then the statement with the
substitution made is also true. If the original statement is false, then
the statement with the substitution made is false.
Recall that A = B means that A and B are names for the same
thing. The substitution property tells us that we may replace A by B in
any statement about A without changing the truth or falsity of the
statement.
2,8 Equality and Number Operations 81
Example Consider these two statements:
(1) 7 = 2 + 5
(2) 3 + 4 = 7
From (1) and (2) it follows by the substitution property of equality
3+4 = 2 + 5.
Reflexive Property of Equality. For any thing a, a = n.
Symmetric Properly of Equality. For any thing a and for any thing
b t if a = h, then h = a.
Transitive Property of Equality. For any things a, b, c, if a = b and
b = c, then a = c
Addition Property of Equality for Numbers. If a, h> c are real num
bers, and if a = h y then
a + c = b + c.
More generally, if a, fc T c, d are numbers such that a = b and c = d*
then
a + c = fr + d and a — c= b — d.
(It is appropriate to include subtraction here since each difference
can be expressed as an addition. Thus a — c — a + (— c). Also if
c = d, then — c = —d.)
Multiplication Property of Equality for Numbers. If a, b, c arc
real numbers and if a = b t then
ac = be.
More generally, if a, b> c t d arc numbers, and if a = /; and c = d,
then
ac
= bd t
and if i
c=£0,d=£Q f then
c
(It is appropriate to include division here since each quotient can
be expressed as a product . Thus — = a* —  Also, if c = d* c =£■ 0,
c c
<f=^0\ then 1 = 1)
c d
82 Separation and Related Concepts Chapter 2
Commutative Property of Addition. If a and b are real numbers,
then
a f b m b 4* a.
Commutative Property of Multiplication. Tf a and h arc real num
bers, then
ab = ha.
Associative Property of Addition. If a, h t c are real numbers, then
(a + h) + c = a + {h + c).
Associative Property of Multiplication, If a z /?, c are real numbers,
then
[ab)0 = a(bc).
Distributive Property of Multiplication over Addition, If a, h, c are
real numbers, then
a(b + c) = ab + ac and [a + b)c = ac 4 fee,
EXERCTSES 2.8
■ In Exercises l20 t name the property that justifies the given statement
L tf x = 5 and x = y. then, y = 5.
2, If ac = 5 and 5 = y, then x = y,
a 7 = 7
4. If AB = CD, then CD = AB.
5. If 3 = x and 4 = y, then 3 + 4 = x 4 y.
6* if 3 = x and 4 = y, then 12 = xy,
7. If § is a coordinate system, then $ = g.
8. If x 4 y = z and if x = a + h, then (a + b) 4 y a ft
9. If lit = tcandif a 4 fr = uandc 4 d 1 = c, then (a 4 fr)(c f tt) = 10.
10. If £ = c, tiien a = c/j.
o
IL Ifa + fc=c, then (o 4 fc) 4 (&) = c + (b).
12. If f";? _ 6 lhcn x _ 5 _ ^ _ Q j
13. 3(20 + 5} = 3 • 20 4 3 ■ 5
14. 3(xy) = (3%
15. (2 4 3)5 = 2 • 5 + 3 • 5
16. (2 4 3){4 4 7) = (2 4 3)4 4 (2 4 3)7
17. (3 + 23} + 7 = 3 4 (23 4 7)
2.9 Sohrfmg Equation* 83
IS. 5 + i=i + 5
19. 3 + (4 + 7) = 3 + (7 + 4)
20. 3{7 + x) = (7 + x)3
2.9 SOLVING EQUATIONS
The properties of equality and number operations are useful in solv
ing elementary equations.
Example 1 Solve for xt x + 3 = 52.
Solution:
x + 3 = 52 Given
x = 49 Addition property (Acid 3)
Example 2 Solve for *; 3,t  5 = 7x + 2.
So?u£ion:
&r  5 = 7x + 2
Given
3:t = 7* + 7
Addition property (Add 5)
4x= 7
Addition property (Add — 7x)
i= I
4
Multiplication properly (Mult
by i)
Example 3 Solve for
x  5 x + 8
10 12
Solutions
r — a _ a: +
10 12
Given
12(x  5) = J0{x + 8) Multiplication properly (Mull.
by 120)
12r — 60 = 10* + 80 Distributive property
12* = lOt + 140 Addition property (Add 60)
2*  140 Addition propei+y (Add — lOx)
* = 70 Multiplication property ('Mult.
by^>
84 Separation and Related Concepts
Chapter 2
Example 4 Solve for x\
X — o
x7
(**T)
Solution:
X5 _ 5
* 7 " 8
8(x  5) = 5(x  7)
fix  40 = 5*  35
3* =5
X =
5
Given
Multiplication property (Mult
by8(*7))
Distributive property
Addition property (Add
(  5x + 40))
Multiplication property
(Mult,
JErompte 5 Solve for « x + ° = * * + 5x.
^ o
Solution;
Jl±A = JLzi1+5x
2 3
3(* + 5) = 2(x  1) + 30x
+ 15 = 2x  2 + 30x
29*= 17
x =
XL
29
Given
Multiplication property (Mult,
by 6}
Distributive property
Addition property (Add
(32a:  15))
Multiplication property (Mult.
by i)
Although the list of properties does not explicitly include the divi
sion and subtraction properties of equably for numbers, they are in
cluded implicitly. For example, dividing by 7 is the same as multiplying
by \, and subtracting 5 is the same as adding —5, In Example 1 you
could think of subtracting 3 and justifying it by the subtraction prop
erty instead of adding — 3 and justifying it by the addition property.
In Example 2 you could obtain the last step from the preceding step by
dividing by —4 and justifying it by the division property instead of
multiplying by — j and justifying it with the multiplication property,
Note, however, that the multiplication property does not permit us
to divide by 0, since is not the reciprocal of any number.
2.9 Solving Equations
Example 6 Solve for x: a(x + x t ) = 5, where a =£ 0.
Solution:
a(x + «J = 5
* + x x  ^
a
x = — _ Xl
Given
Division property (Divide by a)
Subtraction property (Subtract
EXERCISES 2.9
In Exercises 120, solve for x. Name the properties that justify the steps in
your solution. Express your answer in simplest form.
1. 3x + 3 = 2x + 4
2. 3s*  4  4x + 3
3. i* = J*+i
4. 0,75*  100
5. 4(x  64) = 28
i±4£«a
5
 3(*~1) _. 1
' s s
" 85 9 (6)
ft x ~ l  'J ~ 3
* 7  I " 15  3
a:  10 fc  ()
11, 4x  5 = 14c  75
_1_3  _]0(8j
ia x 4 4 = 40  *
*7 „ 7 (3)
14.
10.
11  10
5 1 1
15. 3{x + 15) = 2{x + 15)
16. 3(.v + 15) + 2{ X + 16)
17 '7^ocH^ 100 >
ia ~5— = 7 + '^0^
20. 0.3x 4 0.8x a 220
1 
In Exercises 21 40, solve for x. Express answers in simple form.
21.
3* +
5 BS
22.
3(x
1) + 2(* 
1)
= 4*
23,
5* +
7 = 3{* +
4)
5
24.
x —
2
1 _ x3
3
25.
2ar
3 5x
1
26.
* 1
£***
27. ox 1=4
2S.
g 1
1
x 1
I* ~ ?
5a:— 1
73
x2
3
86 Separation and Related Concept*
30. 5x ~ 1 = ^nl
6*  7 6x  8
31. * + % =3
32. 3x + y =
34. x3 = Jk + 4
35. 5( j  3) = 3{* + 4)
Chapter 2
36. 5*  3 = 3* + 4
37,*~
k + 2
2 3
38. 4x  5 = 3«i + 4xa
x — *i 5
39.
*2 ari
*+! =
7
5
2.10 EQUIVALENT EQUATIONS
In solving an equation, wc find the number (or numbers) that
"satisfies" the equation. Such a number is a root of the equation. In
Example 1 of Section 2.9 the root of the equation
x + 3 = 52
is 49. Note that 49 satisfies the equation since 49 + 3 as 52 is a true
sentence and that no other number satisfies the equation. The set of
all roots of the equation x 4 3 = 52 is
since 49 is the one and only root of the equation, {49} is the solution
set of the equation x 4 3 = 52.
In each example of Section 2.9 we used properties of equality and
number operations to obtain other equations that have the same solu
tion set Equations that have the same solution set are called equivalent
equations. Note in Example 2, for instance, that
anc
3x
5 =
7* + 2
3ac =
7x + 7
4x 
7,
x —
7
4
are four equivalent equations. The solution set of each of them is
HI
Sometimes the properties of equality and number operations are
used to produce equations not equivalent to the given equation.
2.10 Equivalent Equations 87
Example I
x = 5 Given
0=0 Multiplication property (Mult, by 0)
It is true that
if x = 5, then = 0.
Indeed = is a true sentence regardless of what may be true about x.
The solution set of x = 5 is obviously {5}. The solution set of =
is the set of all real numbers. Even though it is true that
if x = 5, then = 0,
it is not true that
x = 5 and = are equivalent equations.
When we went from x = 5 to = 0, we went from an equation with
one root to an equation with infinitely many roots. We did not lose any
roots, but we certainly gained many of them I
Example 2
■■«=T +1
x(x 2J= {* 2) + l(x2)
x*  2* = x 2 + x 2
x 2  Ix = 2x  4
x 2  4x + 4 =
(x  2)2 =
x~2 =
x = 2
In this example we gained a root somewhere along the way. Trie
solution set of x = 2 is obviously {2}> but 2 is not a root of the original
equation. Why? The solution set of the given equation is the null set.
If we try to reverse the steps in this "solution/' we can Justify each step
except the last one. Given x = 2. we cannot justify going from
x{x  2) = (x  2) + l(x  2)
to * = 1 f 1 by dividing by x — 2, The multiplication property
permits us to divide by any number except and x — 2 = if x = 2.
88 Separation and Related Concepts Chapter 2
Example 3
x 2 = 3x
* = 3
In this example we multiplied both sides of x 2 = 3* by — (or divided
both sides by x). In doing this we lost a root. How did this happen? Di
viding by x is legal except if x = 0. Is a root of the given equation?
Is 2 = 3 • 0? Yes, it is. The solution set of the given equation Is {0, 3}.
The solution set of x = 3 is {3}. The equations x 2 = 3* and x = 3
are not equivalent equations.
Tn solving BQ equation through a seqin.:iu.i' <j equations it is advis
able to check each step for a possible loss of roots. If there is a value of
x that might be lost as a root (as is lost in going from x 2 = 3x to x = 3
in Example 3), it should be identified and checked by substitution in
the original equation. Of course, it is a root if and only if it satisfies the
original equation.
To make sure that no roots are gained in the solution process you
may (I) check each root of the final equation by substitution in the orig
inal equation or (2) check to see if the equations can be obtained in re
verse order without a loss of roots at any step. If there is no loss going
backward, there is no gain going forward. If a root is lost going back
ward, it was gained going forward, and hence is not a root of the given
equation. In Example 2 we have a sequence of equations and there is
no loss of roots in going forward. When we reverse the steps, there is
no loss of roots at any step until the last one. 2 is a root of
x(x  2) = x  2 + l(x  2)
x — 2
but not a root of x = + 1<
x — 2
2 is lost as a root going backward; it was gained as a root going forward.
Example 4
1 = I Given
X — i
x — 1 = x  7 Multiply by x — 7
0=0 Subtract x — 7
In this example we have a sequence of three equations and the so
lution set of the last one is the set of all real numbers. In reversing the
2.10 Equivalent Equations 89
steps we do not lose any roots until the last step. To get
x — 7
from x — 7 = x — 7 we divide both sides by x — 7, and this is not
permissible if x s= 7. Checking, we see that 7 is not a root of the given
equation. The solution set of the given equation is the set of all real
numbers except 7.
Example 5 n\ y — 2 _ _
x — 2
(2) y  2 = 7(*  2)
In this example there are two variables, x and y. In Equation (2)
we may wish to think of x as the independent variable and y as the de
pendent variable. A solution f Equation (2) is an ordered pair of num
bers {a, b) such that Equation (2) is satisfied when x is replaced by a
and y is replaced by b. Thus (3, 9) is a solution of Equation (2) since
9  2 = 7(3  2)
is a true sentence.
If we solve Equation (2) for gf, we get
y = 2 + 7(*  2) or y = 7x  12.
Let a be any real number whatsoever. Then x = a and y = la — 12
satisfy Equation (2). For upon substituting a for x and la  12 for y
in (2) we get
(la  12)  2 = 7(0  2),
which is a true sentence. Indeed, the set of all ordered pairs
(a, la — 12), where a is a real number, is the solution set of Equation
(2). Any other letter can be used for the symbol a here. Thus we could
say, if we wish, that the solution set of (2) is the set of all ordered pairs
(x> 7x — 12), where % is real.
Is the set of all ordered pairs fa la  12) also the solution set of
Equation (1)? Let us check Substituting a for x and la  12 for y in
Equation (1), we get
(7a  12)  2 
a 2
or after simplifying, ' ( fl ~ ) = 1. This is a true statement for every a
a — 2
90 Separation and Related Concepts Chapter 2
with one exception. It is not true if a = 2, The solution set of (1) is the
set of all ordered pairs (a, la — 12) for a ^ 2, Equations (1) and (2)
are not equivalent
Some of you have graphed equations like (2). You know that its
graph is a line. What is the graph of (1)? The graph of (1) includes all
the points of the graph of (2) except the point with abscissa 2, that is,
die point (2, 2). The graph of (!) is the union of two opposite halflincs.
Or to use a bit of informal language, a line with a hole in it.
EXERCISES 2.10
In Exercises 120, check by substitution to see if the given value of x is
root of the given equation,
1. X 2 + 5x + 6.25 = 0, x = 1
2. a* + 5x + 6.25 = t x= 2
3. x* 4 5* 4 6.25 = 0, t  2.5
4. 3(x  5) + 7(*  5) = 10(x  5), x = 5
5. 3(x  5) + 7(x  5) = 10(x 5), x = 5.1
6. 3(x  5) + 7(x 5) = 10(x 5),i= 137.3
7 x  2 _ 2x + 3 T _ ,
" 6x~T9*
= 2xJ4 2
St
■ 6
ar —
2
3x
6
x —
2
3x
6
x —
2
3x
6
1
Gx + 9
2*43
fix  9
x= 1.5
10. *=% = ^4 * = 1°°°
Gx + 9
]£ _J_ = _^,*=2
x 4 2 a :  2
x ■+■ 1
14. 1 +
x + 1
1
16.
x + 1
2143
x  3
x —
1
x a 
l
1
2x
X —
1
x a 
l
1
2x
X*
X —
1
1
7
 5
x = 
6
4— l — = ^— t x= 1
2.10 Equivalent Equation* 91
x — 2 * + 3 i4
19.^+*i + *^i=3.x=4
x2 T x + 3 x 4
2a x^ + JL ^3 x4 = _ 3
x — 2 x + 3 x— 4
In Exercises 2130, check to see if the two given equations are equivalent.
If they are not equivalent, state whether there is a loss or gain of roots in
going from the first equation to the second one.
21. x* = 2x, x = 2
22.
2x=3,x
= 1.5
23.
x = 5, x*
= 25
24.
x = 5, x2
= Sx
25,
x = 5, x 
 1 =4
26.
x = 5,2x
■ I =
11
27,
3x + 2x =
: 6x, X
=
28.
x2
3
x 3
4
4(x
2)
29.
x2
3
x  3
x 
x —
2
3 "
3
4 J
1
30
x2
x3
X —
3
4
3
4 '
X —
2 "
3
3(x  3)
In Exercises 3135, find an ordered pair of real numbers that satisfies
second of the given equations but not the first.
22
X
s 5, tj — S s 5x
33. l^i = ^i, 2x  ij  4 =
x — 4 9 — 4 y
^ x2 = 2 x2 =JL 2
92 Separation and Related Concepts Ciarlr „'
■ In Exercises 3645, find the solution set of the given equation.
36. 3x  4* = 8  5
37. %  2) + 7{x  3) = 30
42, M — £ a 2
x— 1
38. x +  + 1 = —
x x
39. ^ = x
x
40. x  1 = x  2
41. 2*  2 = 2(x  1)
43. * ~ * + «
45
+ ^4=3
x1 x2 x3
44. *~? = 5
x2 x2 =2 / x2 \
* x3 x3 \x3^
CHAPTER SUMMARY
There were three BETWKENNESS POSTULATES and three SEP
A RATI ON POSTULATES in this chapter. We list them below by name
only. Try to state each of them in your own words. Draw a picture illustrat
ing what each postulate says.
9, THE ABC BETWEENNESS POSTULATE,
10. THE THREEPOINT BETWEENNESS POSTULATE.
11. THE LINEBUILDING POSTULATE,
12. THE ITXE SEPARATION POSTULATE.
13. THE PLANE SEPARATION POSTULATE.
14. THE SPACE SEPARATION POSTULATE.
The following concepts were defined in this chapter. Be sure that you
know all of them.
SEGMENT
INTERIOR OF A SEGMENT
HAY
INTERIOR OF A RAY
OPPOSITE RATS
ANCLE
INTERIOR OF AN ANGLE
EXTERIOR OF AN ANGLE
TRIANGLE
INTERIOR OF A TRIANGLE
EXTERIOR OF A TRIANGLE
QUADRILATERAL
TWO SIDES OF A POINT
ON A LINE
TWO SIDES OF A LINE
IN A PLANE
TWO SIDES OF A PLANE
IN SPACE
HALFLINE
OPPOSITE HALFUNES
HALFPLANE
OPPOSITE 1IALFPLANES
ILYLFSPACK
OPPOSITE H ALFSPACES
Review Exerciset 93
A set of points is called CONVEX if for every two points P and Q in
the set, the entire segment ¥Q is in the set. The null set and every set of
points that contains just one point are also said to be convex. Each of the
following sets is a convex set: segment, ray, line, plane, h&lflinc, halfplane,
balfspaee, interior of an angle, and interior of a triangle.
Six theorems were stated and proved in this chapter. Study them again
so that you know and understand what they mean.
The last part of this chapter contains a review of elementary algebra.
You should know and be able to use the properties of equality and number
operations that arc useful in solving equations.
REVIEW EXERCISES
In Exercises 115, indicate whether the statement Is true or false.
L If points A, B, C and line t arc in the same plane, and if A and B are on
the opposite sides of J, then C" must be either on the A side of I or on the
Bside of I
2. If B and C are two distinct points on the same side of line n in plane a,
then every point of EC is on the Bside of n.
3. UABC (point B is between points A and C) and BDQ then ADC
and ABD.
4. If RST and QST, then RQS and RQT.
5. If points R and S are on opposite sides of line m in plane a and points
R and T are on opposite sides of m, then S and T are on the same side
of m.
6. The betweenness relations RST and RUT uniquely determine the
order of the points R, S, T t U on a line,
7. If two rays intersect, they have one and only one point in common.
8. AJ9 = M.
9.AB = bX
10. opp A~S = BA
1L AB = BA
12. ab n bX = AB
13. AB U B~X = A~B
14. opp AB D opp B~A = A 9
15. If point B is between points A and C, then BA and BC are opposite rays.
94 Separation and Related Concepts Chapter 2
■ Exercises 1620 refer to Figure 24S.
A
16. How many angles are determined by the segments shown In the figure?
17. How many triangles are determined by the segments shown in the fig
ure? Name them.
18. Point E is in the interior of two angles. Name them.
19. Only one of the labeled points in the figure is in the interior of each of
three different angles. Name the point and name the three angles.
20. Two of Ae labeled points in the figure are not in tlie interior of any of
the angles. Name them.
21. Draw A ABC. Mark a point D such that D is between A and B. Mark
a point E such that B is between C and E.
(a) Docs ED intersect AS? Why?
(b) Does EZ> intersect BC? Why?
(c) Does ED intersect A£? Why?
22. Define the interior of £PQR>
23. Cicen: Line 1 f£ line m.
Point C is between points A and B oa liue L
Point C is between paints D and E on line m.
Point C is between points C and E.
Point F is between points A and E.
(a) Which point (C, G, or F } is In the interior of I ACE?
(b) Indicate whether each of the following is true or false.
(1) Point F is on line I,
(2) Point F is on line iiu
(3) Point F is a point of / ACE.
(4) Point C is a point of L BCF..
(5) Line FG does not intersect AC.
Review Exercise* 95
24. If r f $, t, u are four distinct coplanar rays having a common endpoint
and if no two of the rays arc collinear, how many angles are formed by
these rays?
25. Which of our postulates guarantees that a halfplane is a convex set?
26. Is the intersection of two convex sets always a convex set?
27. Explain why the interior of an angle is a convex set.
28. Is a line with one point deleted a convex set?
29. Is the. union of two convex sets always a convex set?
30. Describe two convex sets whose union is a convex set.
In Exercises 3135, solve for x.
31. 3x  5 = Ix  25
32. 2(r  5} ■ 3x  5
33.^1+1=^
34, 1.75x = 17.50
35. x  1 + 2(x  1) = 17{*  1) + 14
In Exercises 36 40, find the solution
set
36,
x + x =
2*
37.
x + 1 =
x + 2
38.
X
39.
x1
2*2
_ 1
" 2
40,
X  1 =
Iff* 9
Chapter
HeUa Hammld/Bopho GuiUumetttt
Distance
and Coordinate
Systems
3.1 INTRODUCTION
At this point in our formal geometry we have do postulates con
cerning the sizes of objects. We have no basis for saying how big an
object is or even for saying that one object is bigger than another. Al
though the word "size" is often used in informal speech* it is not in the
official vocabulary of formal geometry. Instead we talk about the
length of a segment, the measure of an angle, the area of a rectangle,
the volume of a sphere, the distance between two points, and so on.
In elementary mathematics it is customary to draw a horizontal
number line with numbers increasing from left to right and a vertical
number line with numbers increasing in the upward direction. In some
applications it may be more convenient to order the numbers from
right to left or from up to down. A number line, such as the one in Fig
ure 31 in which the numbers increase from left to right, is a good de
vice for illustrating certain relationships among real numbers. One of
them is the order relation. The fact that 3 is greater than 1 is consistent
With the fact that 3 is to the right of 1 in Figure 31. The fact that 1 is to
the right of —2 agrees with the fact that 1 is greater than —2.
! ! ►
4321 I 2 3 4 5 6 Figure 31
93 Distance and Coordinate Systems Chapter 3
Another relationship is based on the betweenness relation for real
numbers. Since — is between — ■ and %, the point marked — lies
3 5 § 3
between the points marked — and — on the number line shown in Fig*
ure 32.
Figure 32
This chapter is about distance and coordinate systems. We begin by
considering the distance between two points. The idea of a coordinate
system on a line is an extension of our ideas about a number line. Co
ordinate systems are useful in developing the properties of distance.
3.2 DISTANCE
Asking "How long is a certain segment?" is equivalent to asking,
"How far apart are the endpoints of that segment?" In the world of
real objects we can answer the question, "TTow far apart?" by using a
physical ruler. We might determine that two points P and Q are 2 yd.
apart, or 6 ft. apart, or 72 in. apart
To make a physical ruler graduated, say, in inches, we must know
what 1 im is. We must have a segment 1 in. long. In the United Stales
the accepted relation between inches and meters is 39.37 in. =
1 meter. For many years the meter was described officially by two
marks on a phttinumiridium bar kept in France. These marks repre
sented the eildpointS of the segment that was the official meter. Al
though the modern standard for measuring distances is now based on
the wavelength of a certain kind of light, the idea that a given segment
may be a standard or unit for measuring distance is important in our
geometry. Tn the remainder of this section we discuss informally some
of the basic properties of distance and then state these properties for
mally as postulates.
Given any segment, say FQ y we could agree that this segment is the
unit of distance. Thus we might start with a given stick and say, "Let
the distance from one end of this stick to the other end be 1, and let
us call this unit of distance the "stick/' Then the given stick or unit is
the basis of a system of distances which we might call the sticksystem.
In this system the distance between any two different points is a posi
tive number. In particular, the distance between the endpoints of the
3.2 Distance 99
unit stick is 1, Tf the distance between points R and S is 3 in this system,
this means that R and S are 3 times as far apart as the endpoints of the
unit stick. If the distance between points U and Vis 1 in this system,
this means that V and Vare just as far apart as the endpoints of the unit
slick.
Speaking more formally, each segment PQ determines a distance
function. The domain of this function is the set of all segments, or if
you prefer, the set of all pah's of distinct points. The range of this func
tion is a set of positive numbers. It is convenient in developing the con
cept of distance to take the distance between a point and itself to be 0.
Then the domain of a distance function is the set of all pairs (not neces
sarily distinct) of points and the range is a set of noimcgative numbers
including 0. Later, after we adopt the Ruler Postulate, it will be obvious
that the range is the set of all nonnegative numbers including 0.
Our first Distance Postulate is related to the idea that any stick can
serve as a unit of distance. We mi^it develop a formal geometry with
one distance f miction. This would be like using inches for all distances
whether they are thicknesses of paper or distances between stars. Our
first postulate, the Distance Existence Postulate, reveals our preference
far recognizing the possibility of various units of distance i n our formal
geometry. The other Distance Postulates are motivated also by phys
ical experiences with distance, betweenness, and separation.
If A, B, C are three distinct eollincai points with 8 between A and
C, then wc want the distance between A and B plus the distance be
tween B and C to be equal to the distance between A and C as illus
trated in Figure 33.
14*
a ' T c
Figure 33
If A, B, C are three noncollinear points, then we want the distance
from A to B (the same as the distance between A and B) to be less than
the distance from A to C plus the distance from C to B as illustrated in
Figure 34.
Figure 34
100 Distance and Coordinate Systems Chapter 3
We are now ready to State our Basic Distance Postulates in winch
we exercise extreme care in the choice of words so that we agree on
exactly what they mean. As always we need to know precisely what
we are accepting without proof in the building of our formal geometry.
We use "unique" to mean "one and only one" or "exactly one."
Basic Distance Postulates
POSTULATE 15 (Distance Existence Postulate) If AB is any
segment, there is a correspondence which matches with every segment
€D in space a unique positive number, the number matched with AJi
being 1.
If C and D are distinct points, there is a unique positive number
matched with CD according to Postulate ] 5, We may also think of this
Dumber as matched with the set (C, D). If we associate this number
with the segment CD, we think of it as a length. If we associate the
number with the set {C, 1?}, we think of it as a distance. Postulate 15,
however, says nothing about the distance from C to D if C = D; in
other words, it says nothing about the distance between a point and
itself. We take care of this with a definition.
Definition 3.1 The distance between anv point and itself
is a
Definition 3,2
L The correspondence that matches a unique positive num
ber with each pair of distinct points C and D t as in Postu
late 15, and the number with the points C and D if
C = D. as in Definition 3. 1 , is called the distance function
determined by AB or the distance function based on JT&
2. The segment AB that determines a distance function is the
unit segment for that distance function.
3. The number matched with C and D, as in Postulate 15, is
the distance from C to D or the distance between C and D
or the length of CD.
Notation. If P and Q are any points, not necessarily distinct, then
the distance between P and Q in the distance function based on AB is
denoted by FQ{in AB units) or simply by FQ if the unit is understood.
3.2 Distance 101
Example f (Informal) Suppose AB is a segment 1 in. long, CD is a
segment 1 ft. long, and EFis a segment & in. long. Then
EF (in AB units) = 6
and
EF (in W unite) = .
Although Postulates 16 and 17 could be omitted and proved as the
orems after the Ruler Postulate is adopted, they are included in our
formal geometry in order to simplify the development.
POSTULATE 16 (Distance Betweenmss Postulate) Tf A, B, C
are collinear points such that ABC, Lhen for any distance function
we have AB + BC = AC. (See Figure 35.)
a B c
4 • • m ►
Figure 35
Example 2 If ABC, AB  6, BC = 8 P then AC  14.
POSTULATE 17 (Triangfe Inequality Postulate) If A, B, C are
noncollinear points, then for distances in any system we have AB 4
BC > AC (Sec Figure 36.)
Figured
Example 3 If A, B, C are noncollinear points and if BC = 10, then
BA + AC> 10.
Example 4 If the lengths of the sides of a triangle are x, y t z T then
x + y > %, y + z>x, and z+ x> y.
Suppose that FQ is a segment 1 in. long and that 1£S is a segment
1 ft. long. Suppose that A and B are two different points and that C and
102 Distance and Coordinate System'. Chapter 3
D are two different points. We use distances to compare how far A is
from B with how far C is from D, If we say that it is 3 times as far from
A to B as from C to D, we mean that AB  3 CD, or that ^ = 3.
^ CD
(See Figure 37.) Our experience with physical measurements tells us
that we should get the same comparison regardless of which distance
function we use. Thus
.r AB (in PQ units) n . AB (in M units) rt ,
* W (in TQ units) = 3 ' *" CD (in M units) = 3 ^
For example, if AB = 72 (in inches) and CD = 24 (in inches), then
AB = 6 (in feet) and CD = 2 (in feet). Thus
AB (in inches) _ 72 _ AB (in feet) 6
CD (in inches) "" 24 " CD (in feet) " 2 =
This is the basis for our next postulate.
POSTULATE 18 (Dhtance Ratio Postulate) If PQ and RS are
unit segments and A, B t C, D axe points such that A^B.C^A then
AB (in 2^ units) AB (in M imits)
GO (in PQ units) "" CD (in 15 units)
or, equivalently,
AB (in /p units) CD (in PQ units)
AB (in RS units) " " CD (in R5 units) "
In making physical measurements we recognize that there are
many accurate foot rulers. Measurements made with an official foot
ruler and an accurate copy of one should agree. Let us see how Postu
late 18 is concerned with this.
Suppose that PQ and H5 are unit segments and RS (in FQ
units) = 1. This means that the length of RS in PQ units is 1 or, infor
mally, that M is a copy of FQ. If A and B are any points, it follows from
Postulate 18 that
AB (in TQ waits) __ AB (in A S units)
RS (in PQ units) " RS (in M units) "
3.2 Distance 103
Since we assumed that RS (in FQ units) = 1 and since RS (in Jf3
units) = 1 by the Distance Existence Postulate, it follows that
AB (in TQ units) AB (in RS units)
1 1
or AB (in TQ units) = AB (in RS units).
This proves the following theorem.
THEOREM 3 1 If FQ and RE are segments such that the length
of RS in Pp units is 1, then for all points A and B it is true that
AB (in M units) = AB (in PQ units).
KXERCISJiS 3*2
1. If A and B are points and if CD is a segment, which of the following are
necessarily true about the number AH (in CD units)?
(a) It is a real number.
;b) It is a positive number.
(c) It is a nonnegaiive number.
(d) It is an irrational number.
2. If you know that A and B are distinct points and that W> is a segment,
what can you say about the number AB (in RS units)?
3. If you know that ABC, what can you say about the number
AB + EC ,
AC
4. If you know that A, B, C are noncoHinear points, what can you say about
the number AB + BC ?
5. If you know that A, B, C are points, that A # G* and that FQ and H5
are segments, what can you say about the numbers
AB (in FQ units) , AB (in RS units) .
AC (in FQ units) ! AC (in RS units)
Why is it not necessary to say A ^ C in Exercise 4?
6. If you know that A, B, C are noncoUinear points and that PQ and H5
are segments, how does the number
AB (in FQ units) + BC (in FQ units)
\C (in FQ m.iiHi
compare with the number
AB (in H5 units) + BC (an BS units) a
AC (in RS units)
104 Distance and Coordinate Systemi Chapters
7. Let A, B> C, D, E be distinct points ordered on a line I as shown in the
figure, andjttjually spaced so that AB (in AB units) =: BC (in Tfil units)
= CD (ia AB units)_= D£ (in ZB units). Find the following: AB (in
BC units), £C (in BC units), CD (in BC units), and DE (in BC units).
ABODE
4 . . . « . +t
8. Given the situation of Exercise 7, find the distances AB t BC, CD, DE,
all In AC units,
», Given the situation of Exercise 7, And the numbers AB (hi AB units),
AB (in AC units), AB (is Bl? units), and AB (ii i i 7 ! i 1 nits).
10. Civen the situation of Exercise 7, find AD (in A"B units), AD (in BC
units), AD (in AC units), AD (in BE units), and AD (in AE units).
1 1. Given the situation of Exercise 7, copy and complete the following
proof that BD (in JU units) = 1.
Proof: Expressing ail distances in ACT units we have the following:
1. BD — BC + CD 1. Distance Between ness Postulate
CD
2. jg = 1 2. Exercise 7; Distance Ratio Postulate
3. CD = AM 3, Multiplication Property of Equality
4. BD = BC + AB 4. Steps i, 3; Substitution
5. BD = AB + BC 5.
a AC = AB + BC 6. Q]
7, BD = AC 7.
& AC = 1 8. [7]
9. BD = 1 9.
12. In the Distance Ratio Postulate, take PQ —CD and M = JB. Using
this special case of the postulate, prove that
AB (in CD units)  CD (in AB units) = L
(Does this result seem reasonable? Think of AB as a stick 2 ft. long and
CD as a stick 3 ft. long. Then AB (in CD units) = , CD (in AS units) =
1 1
Land* 4 =10
13. Given A, B, C, D such that AB (in C~D units) = % find CD (in AB units).
(Does your answer seem reasonable? The given information means that
AB is — as long as CD. Your answer means that C~D is how many times
as long as AB?)
14. challenge FfioBLEM. Given segments AB, CJ5» EF, prove that
EF (in AB units)  EF (in CD units) • CD (in JF units).
(A simple example is: The number of feet from E to F equals the num
ber of yards from E to F times the number of feet in a yard.)
3.3 Une Coordinate Systems 105
15. {An Infortnal Geometry Exercise.) fat the figure, suppose that the part
of I between F and M is on the edge of a ruler. On this ruler is assigned
to P and 1 to Q. What numbers should be assigned to C, D t E, and \f?
p q CD KM
4 — * ir « — r 9—9 — M
16. (An Informal Geometry Exercise.) Two number lines / and m are
placed parallel to each other as shown in the figure below. What num
bers should be assigned to S and T on IP What numbers should be as
signed to F, Q, and H on m?
A
s
T
I
4
p
—m —
2
Q
•—
V
m
100
3
a 4 1
17. Refer to Exercise 16. If A on if is directly alxrve B on m and if 100 is as
signed to A, what numl>er is assigned to B?
18. challenge problem. Hefer to Exercise 1 6. If a number x is assigned
to a point on land if the number x' is assigned to the point directly be
low it on m f express x' in terms of x.
19* In the Distance Hah'o Postulate we stated that the two equations were
equivalent, (a) Derive the second equation from the first one using some
of the properties of equality, (b) Derive the first equation from the sec
ond one using some of the properties of equality.
3.3 LINE COORDINATE SYSTEMS
A key feature of a number line is that different points are matched
with different numbers. In fact, we consider al the points of the line as
matched with all the real numbers. Such a matching is called a oneto
one correspondence between the set of all points on the line and the
set of all real numbers.
Another key feature of a number line is that the distance between
any two points is the absolute value of the difference of the numbers
matched with those two points. A unit segment for these distances is
the segment whose endpoints arc matched with and 1.
Example '" Figure 38,
AB= 11 (2) = 1. J BC=0(l) = l = l.
CE = 2  0 = 2. AC = 0  (2) = [2 = 2.
BD = 1  (1) = 2. BE = 2  (1) = 3 = 3.
AE= 2  (2)1, = 4. AE = 2  <2) = 4 = 4,
A B C D S
106 Distance and Coordinate Systems
Chapter 3
The concept in formal geometry that corresponds to a number line
in informal geometry is the concept of a coordinate system on a line.
Coordinate systems are useful tools in our formal development of ge
ometry for planes and spaces as well as for lines.
Definition 3,3 Let FQ be a unit segment and / a line. A
coordinate system on I relative to PQ is a onetoone cor
respondence between the set of all points of I and the set of
all real numbers such that if points A. B, Care matched with
the real numbers a* b t c, respectively, then
1. B is between A and C if and only if b is between a and c and
2, AB (in TQ units) = \a  b\.
Definition 3,4
2* The origin of a coordinate system on a line is the point
matched with 0,
2. The unit point is the point matched with 1.
3. The number matched with a point is its coordinate.
These definitions describe and help us to think about a coordinate
system. But a very important issue must not be overlooked here. How
do we know there are things such as coordinate systems in our formal
geometry? To answer this question we return to our experiences with
number lines and physical measurements.
Let us suppose that TQisa. unit segment, say a segment 1 cm, long,
and that A and B are any two distinct points on I. From our experience
with physical rulers we know that we can lay a ruler graduated in centi
meters alongside I and measure distances starring from A and extend
ing toward B as shown in Figure 39.
♦I
Ffaove 3D
l' entire cere
Of course, we can just as well start at B and measure toward A as
shown in Figure 310, These ideas suggest our next postulate.
*/
Figure 310
Centimeters
3,4 Rays, Segments, and Coordinate* 107
POSTULATE 19 (Ruler Postulate) If AB is a unit segment and
if P and Q are distinct points on a line l t then there is a unique coordi
nate system on / relative to AB such that the origin is P and the coordi
nate q of Q is ft positive number. (See Figure 311.)
I
q(q>0} Figure 311
We are now ready for a theorem.
THEOREM 3,2 (The Origin and Unit Point Theorem) If P and
Q are any two distinct points, then there is a unique coordinate
system on PQ with P as origin and Q as unit point.
Proof:
1. There is a unique coordinate system 1. Ruler Postulate
on PQ relative to TQ t with the coor
dinate of P equal to and the coor
dinate q of Q a positive number.
2. PQ (in Pp units)  q  = q. 2, [T]
3. PQ {in Pt> units) si 3. [7]
4 q = 2 4
5. Q is the unit point. 5, [7]
3,4 RAYS, SEGMENTS, AND COORDINATES
I^t A and B be two distinct points on a line I. We know from the
Origin and Unit Point Theorem that there is a coordinate system $
with A as origin and R as unit point. Let X be any point of I and let x be
its coordinate in %, Then it follows from the definition of a coordinate
system that A is between X and B if and only if is between x and lj
X is between A and jB if and only if x is between and I ; B is between
A and X if and only if 1 is between and x. (See Figure 312,)
X
A
a
X
A
X
s
1
A
JC
1
B
X
I
X
Figure 312
108 Distance and Coordinate Systems Chapter 3
Now is between x and 1 if and only if t < 0; x is between and 1
if and only if < % < 1; and ! is between and x if and only if .* > 1 .
It follows from the definitions of ray, ray opposite to a rav, and segment
that:
ojtpAB = {X :x<0) 4 2 A i >t
AE = {X : < x < 1} 4 ^ — G
A$ = {X : x > 0} 4 4 — $
►/
1
= {X i x < I ) < d — £
+f
■w
Similarly, if C and D are two distinct points on a line I with coordi
nates 2 and 5, respectively, and if X is a (variable) point on / with co
ordinate x f then:
<5B = {X : 2 < x < 5}
a {X : i > 2}
= {X:x< 5}
opp CD a (JE : x £4}
opp D? = {X : i > 5}
CD = {X : .r is real}
.'
£_
%
2
5
rl
_s
9
« — '
2
5
w 1
6
V
2
3
tl
4 —
—A
23
2
5
C
9
— • ,
2
5
r t
£
Q
* i
More generally, if the coordinates of two distinct points A and B
on a line I are a and b, respectively, then it is convenient to consider
two cases in expressing subsets of AB using setbuilder symbols as
follows.
If <i < b If a > fr
AB = {X;a<x<b) IB = {X : b < % < a)
AB= {X:x> a) AB = {X : x < a]
BX = {X : x < b) ~BA = [X:x>b]
oppA$={X:x<a) oppA& = {X : x > a}
oppBA = (X :x>b] oppBA={Xix<b)
3.4 Rays, Segment*, and Coordinates 109
EXERCISES 3.4
In Exercises J 5, a line / with a coordinate system $ is given. The coordi
nates of points A,B t C arc — 3 t 0, 4, respectively. In each exercise draw the
graph of the set and express it in terms of coordinates using a setbuilder
symbol Exercise I has been completed as i sample.
1. BC
Solutions BC = {X : x > 0}
4 fc
3
+ 1
2. BA
3. AC
4. m
5. £2
In Exercises 610, a line / and a coordinate system § are given. In each
exercise, given the coordinates of two points, find the coordinate of a third
point Exercise 6 has been worked as a sample. {Note: cd A means "coordi
nate of A.'*)
a cd A = 2, cd B = 5, cd X = x. If X £ ~&& and AX = 2 ■ AS, find x.
Solution x < 2, AX = 2  x, AB = 3. %  x = 6, x = 4
I i ft.
N
7. cd C = U cd D = 1, cd P = p. If P € c3 and ■— = ^, find p.
8. cdE = 5, cd F = G,cdQ = qAfQ£ oppFE and EF = FQ, findq.
9. cd C = 29, cd H = 129, cd I = i. If GI = III, find L
10, cd J = 15. cd K = Q> cd R = rAf JR = 2 RK T find the two possible
values of r.
In Exercises 1115, given the coordinates of two points on a line, find the
length of the segment joining these two points,
U. cd A m 5, cd B = 173
12. cdC= ~5,cdB= 173
13. cdC = & cdD =  173
14. cd E = 147,5, cd F = 237.6
15. cd G = 19j, cd f* =  17
1X0 Dictince and Coordinate Systems Chapter 3
■ In Exercises 1620, given the coordinates of two points on a line, find the
coordinate p of a third point P satisfying the stated condition.
16. cd A = 5, cd B = 10, P € Ail and AP = 5
17. cdA = o,cdB = 10, P € oppBA and AP = 5
18. «f A = 5. erf B = 10, P € BA and AP = 5
Id. erf A = 5, erf B = 10, P € opn A^ and AP = 5
20. cdC= 10, cdD = Q, P£ CD and PC = 17
■ In Exercises 2125, given a line I and a coordinate system on it in which x
is the coordinate of a variable point X, draw a sketch showing / and the given
subset of/.
21. {X : ~6<x< 2}
22. {X :x< 6}
23. {X : x > 2}
24. {Xix< 6orz>; 2}
25. {X\x< 10}
■ In Exercises 2635, given a tine with points and coordinates as marked in
Figure 313, find the given distance.
CQ D
E
F Q
H
P /
JS X L MT N
3 t 2
I
1
2
f 3
4^6 G 7*8
3 V2 6 /£
V
fl
Figure 343
26. CE
31. /\)
27. DC
32. st
28. HP
33. SF
29. IC
34. A/g
30, FQ
3d. PS
L0
In Exercises 3645, given a line and a coordinate system in which
cJ A = 12, erf B =  8 f cd C = 0, erf D = 4.2, and erf E = \M deter
mine if the given between ijcss relation is true or if it is false.
36. ABC 41. BEA
37. ABD 42. DCA
38. DCE 43. DBA
39. CJ5A 44 BED
40. £CB 45. CAB
35 Segment! and Ccmgrusnc* 111
In Exercises 4850, given that AS is a segment I ft long and that CD is a
segment 1 yd long, find the given distance.
46. EF (in AS units), if EF (in CD unite) = 5
47. CH (in W units), if GH (in AB units) = 5
48. CD (in CD units)
49. CD {in Al units)
50. AB (in (715 units)
51. Copy and complete the proof of Theorem 3.2.
3.5 SEGMENTS AND CONGRUENCE
In Chapter 5 we develop in considerable detail an idea called the
congruence idea. Informally speaking, two figures arc congruent if
they have the "same size and shape." The terms "size** and "shape"
are not considered as a part of our formal geometry. In elementary
geometry, we develop the concept of congruence for segments, for
angles, and for triangles. The concept of congruence for segments is
easy and is appropriate to include here. Intuitively, we feel that all seg
ments have the same shape; hence, they have the same size and shape
if they have the same length. We make this formal in the following
definition.
Definition 3.5 Two segments (distinct or not) are congru
ent if and only if they have the same length. If two segments
are congruent, we say that each of them is congruent to the
other one and wc refer to them as congruent segments.
It is convenient to have a special symbol for congruence; thus
AB s CD means AB is congruent to (35,
It may be helpful to compare the words "congruent and congru
ence" with the words "equal and equality."
7=3 + 4 may he read as "7 is equal to 3 plus 4"
7 — 3 f 4 is an example of an equality.
ABs UD may be read as "AB is congruent to (?D"
AB as CD is an example of a congruence,
When working with segments it is important to note carefully the
difference between equality and congruence. The statement AB = CD
means that A"B and UD are the same set of points, that is, that "AT?"
112 Distance end Coordinate Systems Chapter 3
and "CD" are different names for the same segment. This statement
is true if and only if A = C and B = D, or A = D and B = C. (See
Figure 314.)
• • Or • •
C D D C
Figure 314
The statement AB at UD means that AB and CD have the same
length, that is, AB = CD, Therefore, if AB at W f then AB = CD; and
if AB = CD, then_ AB at CD. Notethat AB = CD implies 33 « CD
but that XB at CD does not imply AR = CD.
We have emphasized the difference between congruence and
equality as these concepts apply to segments. We note next some
similarities. Recall the equivalence properties of equality, the reflexive,
symmetric, and transitive properties. (See Section 2.8.) Congruence
for segments has the same properties, as stated in the following
theorem.
THEOREM 3.3 Congruence for segments is reflexive, symmetric ,
and transitive.
Proof; Let AB be any segment Then its length AB is a number, and
AB = AB by the reflexive property of equality. But AB = AB implies
that ~KB s AR Therefore congruence for segments is reflexive. (The
remainder of this proof is assigned as an exercise.)
Our next theorem expresses formally a simple idea concerned with
adding or subtracting lengths. The theorem expresses this idea formally
in terms of congruences. Theorem 3.4 is followed by Corollary 3,4.1. A
corollary i$ a theorem associated with another theorem from which it
follows rather easily.
THEOREM &4 (The LengthAddition Theorem for Segments)
If distinct points B and C are between points A and D and if
A3 as CD, then A€ss 3D.
Proof: There arc two possibilities as suggested in Figure 315.
i— g g g a SB a
Figure 315
3.5 Segmmts and Cortgru«rtoe 113
Case L If ABC, then RCD, and it follows from the Distance Be
twecnness Postulate that
AB + BC = AC and BC + CD = BD,
Since AS = CD (hypothesis) and since BC = BC (Why?), it follows
from the addition property of equality that
But
+ BC=CD + BC.
CD + BC = BC + CD, (Why?)
Therefore
AB + BC= BC+ CD (Why?)
and AC = BD. This proves that AC ss ED,
Case 2, If A GB, then CBD, and it follows from the Distance Be
tweenness Postulate that
AC + Cfl = AB and CB + BD = CD.
(The remainder of this proof is assigned as an exercise. Note in Case I
that the idea, or strategy, of the proof is the addition of lengths. The
proof of Case 2 involves the subtraction of lengths.)
COROLLARY 3,4,1 If distinct points B and C are between
points A and D and if AC == BD, then AB =ss £J5.
/too/* If B and C are between A and D, then C and B are between A
and P. The corollary follows immediately from Theorem 3.4 by inter
changing B and C that is, by renaming point B as point C and renam
ing point C as point B,
COROLLARY 3.45 If A, B, C, D, £, £_are points such that
AjBC DEb\ AB =s 73E, BC ssEF, then AC s 35F.
iVoo/ Assigned as an exercise.
COROLLARY 3A.3 If A, B s C, D, E t F are points such that
ABC, Z>£i? AB == 0E, A^s Z5F, then BC  £F.
Bkx>/; Assigned as an exercise,
114 Distance a raJ Coordinate Systems Chapter 3
We now raise a question that leads to the final theorem of ibis sec
tion. Suppose that a distance function determined by PQ, say, and a
ray AB are given. Is there a point C on AB such that AC = 7? Is there
only one such point? We know from the Ruler Postulate thai there is
a unique coordinate system on AB relative to FQ such lhat the origin
is A and the coordinate b of B is a positive number. Then
a3= {X:x>0).
(See Figure 346.)
J L
b *
Figure SIB
This means that there is a onetoone correspondence between the
set of all points of AB and the set of all nonncgative numbers. There
fore there is exactly one point C on AB such that Lhe coordinate of C
is Z Then
AC = 7  = 7.
If D is any other point of AB, then the coordinate d of D is not 7 and
AD = d  ^ 7, Therefore there is one and only one point C on AB
such diat AC = 7.
We know from our experience with rulers that we can start at any
point on a line and 'lay off' in either direction (along either of the two
rays on the line that have the starting point as its endpoint) a segment
of any desired length. The example we discussed shows that in our for
mal geometry, given any ray, we can 'lay off," or "construct," a seg
ment which Is 7 units long such that the endpoint of the my is one of
die endpoints of the segment. Of course, we could start with any posi
tive number other than 7 and the same reasoning would apply. We
could start also with a given segment and "lay oF* on AB a segment
whose length is the length of the given segment, thai is, a segment con
gruent to the given segment. The "lay ofF" or "construct" language is
informal All we are really saying is that there is such a segment and
that it is unique* These ideas lead to our next theorem.
THEOREM 3.5 (Segment Construction Theorem) Given a seg
ment CD and a ray AB, there is exactly one point P on AB such lhat
APssCD.
3.5 Segment* and Congruence 115
Proof: Given a ray A3, a segment CD with CD = p t and a distance
function » there is a unique coordinate system § on AB such that A is
the origin and such that the coordinate h of B is a positive number. (See
Figure 317.) Then
"*= iX:x>Q\,
4 =_£
b P
Figure 317
Let P be the unique point of AB such that the coordinate of P is p. Then
AP = p — = p. If Q is any point of AS other than P t then its coor
dinate q is different from p and
= <j0=q=£p.
Therefore there is only one P on A3 such that AP = p, that is, such that
EXERCISES 3,5
In Exercises I~10 t a line I with a coordinate system $ is given. The coordi
nates of points A, B, C, D are —5, 3, 5 t 13, respectively. (Sec Figure 318.)
In each exercise, determine whether the given statement is true or false.
A
B C
D
5
3 5
* r
13
figure 31
I. "KB ss CD
6. AB=sB(7
I.BCs^W
7. AB= lA
3. ATszW
8, JB= £D
4. AB + BC =
AC
9. AC nlE5 =
riJC
,iASUl? =
JTC
10. AC + HD =
: AB + CD
+ 2BC
Iii Exercises 1 1—15 name the property of congruence for segments which
justifies the given statement.
1L If AB at UD t then ITS a ATI
12. If MssUDimdrBsiDE* thenAEsDE
13. FQ as £P
14. If^^^BU^CD ( andOTEF. thenAB^EF.
15. If XY is a segment, then XT S£ XY.
ilfi Distance and Coordfnate System t Chapter 3
16, Complete the proof of Theorem 3.3,
17. Complete the proof of Theorem 3.4.
IS. Draw an appropriate figure and prove Corollary 3,4.2.
10. Draw an appropriate figure and prove Corollary 3.4.3.
■ In Exercises 2025, the coordinates of points B and C in a coordinate sys
— ■
tern on HC are given. If ABC and ~KE ss BC, find the coordinate of A.
20. cdB = 0,cdC=5 23, cdB = $, cdC = $
21. cd B=5,cdC= 24. cd B = 159, cd C = 156
22. cd B= ^234, cd C =  108 25, cd B = 27, cd C = 102
3.6 TWO COORDINATE SYSTEMS ON A LINE
Let A and B be the origin and the unit point, respectively, of a co
ordinate system § on a line /, What is the coordinate of the point P if P
is between A and B and twothirds of the way from A to B? (See Figure
319.) Obviously, adf =  since AP = § and AS = 1.
* • ■ * M
Figure 319 I
Let C and D be points with coordinates 35.0 and 39,8, respectively,
on a line /. What is the coordinate q of point Q if it is between C and D
and twothirds of the way from C to D? (See Figure 320.)
G q D
4 — » + 9—tt
35.0 ? 39B
Figure 320 * 1
We know that there is another coordinate system on / in which the
coordinate of C is and the coordinate of D is 1. If Q is thought of as a
variable point on this line with q as its coordinate in one system and %
as its coordinate in the other system, then the particular point Q we
want lias j as its scoordinate. In this section, we learn how to express
the ^coordinates in terms of the ^coordinates. In this example, the re
lation is
q = 35.0 + 4.8*.
Substituting § for x we get
q = 35.0 + (4.8)() = 35.0 + 3.2 = 38.2,
Therefore the point that is twothirds of the way from the point with
coordinate 35.0 to the point with coordinate 39.8 is the point with co
ordinate 38,2.
3.6 Two Coordinate Systems on a Line 117
The relationship between two coordinate systems on a line is useful
in solving exercises involving points of division like the "twothirds of
die way from C to D" exercise as well as some exercises appearing later
in this chapter. This relationship is also useful in gaining new algebraic
insights. In studying equations like
1/ — 2 f 3x and
Jt
2 1
it is sometimes helpful to think of them geometrically in terms of co
ordinates on a line.
In later chapters Wfl consider coordinate systems in a plane and in
space. In studying a line in a plane or in space you will find it helpful
to think of several coordinate systems associated with the line. What
we do later is a natural extension of the groundwork we are laying in
this chapter.
We begin with a lemma, a "little theorem," that is useful in proving
a "big theorem." Lemma 3.6. J plays a key role in the proof of Theorem
3.6, which is followed by Corollary 3.6, .1.
LEMMA 3.6.1 Let Xi and x 2 be the coordinates of distinct points
Xi and AV respectively, on a line /. If x is the coordinate of a point
X on /, then
AXi X — X\
if X<= XiX 2
A 2 A a % 2 — Xi
and
if X £ opp A'jAV
Proof: First, suppose that X £ XiXi; then either xi < x;? or x 3 < *i
(See Figure 321.) If x t < xg, then xi < x, XX a = x  x u XgXj =
*2 — *i, and
AXi x — Xi
X2X1
X 2
—
Xl
XX,
X
 *1
X2X1
X2
— x L
■>
X%Xi X2 — x%
Xi Xz X X X 3 Xi
• • • M 4 • • • hi
X\ < 12 JC2 < X\
Figure 321
If *a < xi, then x < xi, XXi = x\  x t X 2 Xi = Xi  x 2 » and
XX 1 X\ — X X — Xi
X2X1 l!  X 2 X 2  Xi
118 Distance and Coordinate QyitNM Chapter 3
Next suppose that X £ opp XiX 2 ; then either x x < x 2 or x 2 < x\.
(See Figure 322.) If Xi < x 2j then t\ > x t XX\ = X] — x,
X 2 Xi s %2 — Xu ^d
\ A: 3Ci — x x — *i
X 2 Xi *2 — *i % — a& '
* • • • ► / 4 • • • ►i
x 11 ar 2 a 3 *! j
x 1 < *a *a < *i
Figure 322
If xa < xi, then r > x lt XXi = x — x 1? X 2 X t = x% — x 2 , and
XXi x — Xj x — Xi
X 2 Xi Xi  Xa x 2 — X! '
Therefore, if X £ SjXj, then
XXi x Xi
XgXj X2 — X]
regardless of whether Xi < *2 or x T > x 2 . Also, if X € opp XiX 2 , then
XX 1 x — xi
XjjXi x 2 — Xi
regardless of whether xi < Xa or xi >• X2.
THEOREM 3.6 (The Two Coordinate Systems Theorem) If Xx
and X 2 are two distinct points of a line I, if the coordinates of Xi
and X 2 are *i and x 2 , respectively, in a coordinate system S, andx^
and x 2i respectively, in a coordinate system §*, then for every point
X on /, it is true that
x — %\ x / — xj
Xt — Xi x 2 — x t
where % and x' are the coordinates of X in $ and in §', respectively.
Proof: Suppose that we are given a line / and points X x and X 2 on I
with coordinates Xi and x 2 in § and coordinates xi and x 2 in §' as in the
statement of the theorem. (Sec Figure 323.)
* • • 7ZI ►*
Figure 323 x\ x' t {,$'}
3.6 Two Coordinate Systems on a Line 119
Suppose that X" £ X1X2; then it follows from Lemma 3.6.1 that
XXj x xi XXi xf  x^ x~ ^1 x 1  xj
X2Xi x 2 — xi ' X^Xi x 2 — x{ * x 2 — xi x^ — x[
*2
—
X?
—
*J
*t
—
*f
x 1
«J
Suppose that X £ opp XiX 2 ; then it follows from Lemma 3.6. 1 that
XXj * *i
X 2 X t
XXi
x — Xi
*2 — *1 X' 2 — X[
Therefore for every X on I it is true that
x — x\ x 1 — x[
X'i — X\ x 2 — x\
and the proof is complete.
Theorem 3.6 is closely related to the Distance Ratio Postulate. Ac
V X
cording to this postulate the ratio M s independent of the distance
XzXi
function. Tf we use the distance ftmction of the coordinate system g in
Theorem 3.6 y we have
XX j x — X\\
X2X1 \*2  Xi\
If we use the distance function of the coordinate system §' in Theorem
3,6, we have
XXi = K  «;i
x 2 x, 14^1'
Therefore it follows directly from the Distance Ratio Postulate that
I x — Xi I I x? — xj I
I Ig — *1 I I * 2 — X[ \
It follows from the Two Coordinate Systems Theorem that the equa
tion obtained from this one by omitting the absolute value symbols is
also true.
120 Distance and Coordinate Systems
Chapter 3
COROLLARY 3. 6.1 Let Xi and X 2 be the origin and unit point,
respectively, in a coordinate system §»&. on a line L Let xi and % be
the coordinates of X t and X 2j respectively, in a coordinate system
§ r on /. Let k and x be the coordinates of a point X on / in the sys
tems %k and $ x , respectively. Then
(i)
and
W
x — X\
X2 — Xi
a k t that is, x = xi + k(%2 — *i)
Proo/' (See Figure 324.) It follows from Lemma 3.6.1 that
XX, k
or
But
XbX l 1 
XXi kQ
X2X1 = I 
= *
s *.
XX
XjX
L > a Why?
Therefore
and (!) is proved.
K%Ki
1 *'
x a
x
1
1
(£*) r
*1
X2
*
CgJ
Figure 324
It follows from Theorem 36 that
x 
 xi k 
0
Xi  xi 1 —
3,6 Two Coordinate Systems on a Line 121
Hence it follows that
X2 — Xi
X = *i + k(x 2  X&
and so (2) is proved.
Example / (See Figure 325.) Think of A as X h B as X 2 . Then Xi = h
X2 = 7, Xj = 3, *2 = 15, Then it follows from the Two Coordinate
Systems Theorem that
x 1 x* 3
71 153
for every point X on AB. This equation may be solved for x in terms of
X* or for x* in terms of x. The resulting equations are
S? ss §1? — 4 and x* = 2x + 1.
* • • ♦ *
S 15 ^ (S'j
Figure 335
These equations are useful in finding r' if you know x or in finding x if
you know x\ If you are given that x' = — 6, you can use the first of
these equations to get x = — 3j. If you are given x = 2, yon can use
the second of these equations to getx' = 5,
Example 2 (See Figure 320.) Think of P as X L , Q as X 2 . Then
ar*0 x{7) , , I. , 
4 » ■ ■* ►
i7 3 (£)
Figure 328
(Or we may think of Q as Xi and P as X2. Then the theorem gives
* 5
05
equation obtained by the first method.)
us £ 2. — * ^ , which simplifies to x' = \(x + 7), the same
122 Distance and Coordinate System* Chapter 3
Example 3 Given three coordinate systems with points and coordi
nates as marked in Figure 327, express x in terms of Jfc. Then express x'
in terms of k and, finally, express x' in terms of *.
ABC
■ • • • — ►
a s x
o i ft ($')
Figure 327 n 3 ,' (§)
Solution:
(1) i=4 * ^4 or * = ^ + 2
w 62 10
(2) £  ~ n = * ~ ° or x 1 = 8k + 1 1
6
 2
1 

x' 
11
Jt

3
11
1 

x* 
11
_
x 
 2
(3) sfu =tr or ^^zt+is
In working an example like this one yon may need to write more
details than we have shown. A more complete version of (3), for exam
ple, might he as follows:
3
 11
62
x* 
 11
8
x2
4
x'
11 =
%~
2)
x* =
_2x +4+11
X* 
2ac +
15
Example 4 Given two coordinate systems with points and coordi
nates as marked in Figure 328, express x' in terms of x.
4
A B
X
Future 3.2K
3 I
a 10
IS)
<£")
Solutinu:
rf  (8)
10 (_8)
3^ + 8
10 + 8
x' + S
x 3
13
x 3
4
x3
4
3,6 Two Coordinate Systems on a Line 123
^_U_!9 or *=»">
2
tSxantpte 5 Given two coordinate systems with points and coordi
nates as marked in Figure 329, find x.
B x
• • != ►
10 19 (S>
23 88 x (§') Figure 3S9
Solution:
*{23) 190
38  (23) 100
x + 23 _ 19
38 + 23 B ' 10
*+23 _ t fi
x + 23 = 28.5
* = 51.5
llxamph (i A train traveled at a uniform speed on a trip from Chicago
to New Orleans. If it was 180 miles from Chicago at 7:00 p.m. and 320
miles from Chicago at 9:00 p.m., how far from Chicago was it at
10:20 p.m..'
Solution! (See Figure 330.) Think of hours past noon as forming one
coordinate system and miles from Chicago another one. Suppose the
train is x miles from Chicago at 10j hours past noon, that is, at
10:20 PM. Then
x 180 107 x 180 3^
or
320  180 97 140 2
so
x  180 = 140(~) = i^, x = 180 + 292 = 413^.
6 6 3
1 1 g* (8) i
180 320 x Figure 330
"Hierefore the train is 413^ miles from Chicago at 10:20 p.m.
124 Distance and Coordinate Syittm* Chapter 3
Example 7 Given that 7.  12, p are the respective coordinates of
points A, B t P on a line/, that P C AB, and that AP a 3 AB, find p.
Solution: (See Fjgu re 33 1 .) I n the coordinate system with A as origin
and B as unit point the coordinate of P is 3, Then
4 • • • hi
p 12 7
Figure 331  10
Ezample 5 Given that 7,  12, p are the respective coordinates of
points A, B, Pon a line I, that P £ opp AB, and that AP = 3 ■ AB t find p.
Solution: (See Figure 332.) In the coordinate system with A as origin
and B as unit point the coordinate of P is —3. Then
p  (  12) —31 p + 12
7(uS = ori i§ = 4  p + 12 = 7 «. p = 6 *
< — * — i £ M
12 7 p
Figure 332 i o 4
EXERCISES 3.6
■ In Exercises 15, a, h, c are the respective coordinates of points A, B, C on a
line /. State which point, A, B, or C, is between the other two,
1. a = 0, b = 5, c = 100
2. a = a & = 5, c =  100
3. a= 3,fc = 7,c = 7
4. a = 0.500. & = 0.O50, c = 0.005
5. a = J, h = i c =  j
■ In Exercises 015. a, fc, c are the respective coordinates of points A , B, C on
a line /. From the given information determine whether b < c or b > &
0, AJBC, o = 10, /> = 20 11. ACB, a = 8 t b = Q
7. A~#£ a = 20, b = 10 12. BCA, a a 8, b =
8. BAC, a = 10, 6 = 20 13. lM^, a =  J, 6 = 0.66
9. BAC, a = 20, h = 10 14. BCA, a = §> b = 0.66
10. ACB, a = 0, b = 8 15, BCA, a m , h =
3.6 Two Coordinate Systems on a Line 125
In Exercises 1620, a, b t c are the respective coordinates of points A, B, C
on a line I. From the given information determine the number c.
16. ABQ a = 0,fc = l,AT5s*BC
17. ACB, a = 0, b = hW s*CB
IS. BAC, a  0, b = 1,AH sAC
19. ABC, a = 17, b = B,XB==B€
2ft ABC, a = 17, fc = 8, M m ~BC
In Exercises 2125, there is a sketch of a line with some points and coordi
nates marked. Find the coordinate x.
ABC ,, R Q X
21. i » • • ► 24. 4 • •— « ►
I 1 m 3 13 15 (S)
2*3 (§•) 1 x (S r )
X P Q A B P
22. 4] • • • ► 25. 4 • • • ►
x 1 (£) 16 8 30 (g)
29 B 10 (g'J 1 8 i (g'J
23. 4 *■
3 11 13 <$)
i 1 (S)
In Exercises 2630, there is a sketch of a line with some points and coordi
nates marked. In each exercise write an equation relating x and x r and sim
plify it to the form x f = ax ■+■ h, (Note that a must be a number different
from 0. For if a = 0, then x* = b and every point X would be matched with
the same number in the system §'. Since §' is a coordinate system, we know
tiiat different points must be matched with different numbers.) Check your
answers by substituting the values of x at A and B in the equation to see if
you get the corresponding values of x 1 at A and B.
26. 4
27.
2S,
A
I
X
29.
A
X
B
2
8
X
x'
fin
4
X
x'
4
i
A
B
34%
A
B
X
X
x'
I
&
5
7
i» ■
40
400
X
x'
(S)
A
X
B
~3
9
X
X*
10
IB
126 Distance and Coordinate Systems Chapter 3
■ In Exercises 3135, a subset S of a tine / is given in setbuilder notation.
Sketch the graph of the set and mark the ^coordinates and ^coordinates
of three points of 5, Exercise 3 1 has been worked as a sample.
3L $= {X:x=6fc + 2 l *£2}
Sftlution
ion: S = {x:^2 = jt,fc<2j
*
£
1
X
14
8 2
32, S ss {X : x = 3k + 6, < k < 1}
33, S = {X : x = 3k + B, 0<k< 1}
34, S = {X : .*= 3* 8, * > 0}
35. S = (X :
3C — 3
 7  :
k0
I 0
2
14
+ t
, k is any real number}
In Exercises 3640, A and B are points on a line with coordinates 7 and 12,
respectively. Enid the coordinate of the point F subject to the given
condition,
36. F C AS and AP = 3 • AB
37. F £ AB and AP = j • A23
38. P € opp AB and AP = 3 ■ AB
30, P C AB andgi s 2 (two possibilities)
FB
40, F e BA and AF = ±PB (two possibilities)
In Exercises 4150, A r B,P are points on a line t with coordinates 0, 1, Jt, re
spectively, in a system § and with coordinates —3, 7, x, respectively, in a
system §'. Figure 333 is an appropriate one for Exercise 41.
Figure 333
AT
a
i (S)
7 <£')
"M
41. II P £ AB and ^ = % find fc and *.
AB
42. If F £ opp AB and— a . find fe and k
43. If It a 2. find
AP
AB
P £ AB, or is F £ opp AB?
3.6 Two Coordinate Systems oci a Line 127
44, lfk = 5, find 41 1* P € AB, or is P £ opp AG?
AS
45. Ifx = 17, find A and 44
40. Ifx=27. find ^c and 4^
47. lfx = 27, find^.
48. If* = 37, End^.
49. lf* = 47, find^l.
50. Hx = 0,find^,
51. Given the situation of Example 6 on page 123, express die number x
of miles from Chicago in terms of the number t of hours past noon of
the day the trip began.
52. Given the situation of Example 6, find the "departure" time at Chicago.
53. On a car trip across the country Mr. X stopped at Ridgeville and "filled
it up"; his odometer reading was 35378. Sometime later the gasoline
gauge read  full and the odometer read 35513. Assuming that there is
a constant "gasoline mileage," what does the odometer read when the
gauge reads g?
54. Given the situation of Exercise 53, express the odometer reading m in
terms of the amount x of gasoline in die tank, where x = 1 when the
tank is full and x = when the tank is empty.
55. Mr. X reeendy completed a 180day weight reduction program. If he
weighed 200 lb. on the tenth day and 180 lb. on the 100th day, how
much did he weigh on the 150th day? Assume thai he loses the same
weight each day.
56. Given the situation of Exercise 55, express Mr, X's weight w [hi pounds)
in terms of the number n of days, where n = 1 means the first day of
the program, » = 2 means the second day of the program, and so on.
57. Coder certain standard conditions the freezing point of water is 0°
Centigrade and 32° Fahrenheit; the boiling point of water is 100* Cen
tigrade and 212" Fahrenheit. What is the temperature in degrees Cen
tigrade when the Fahrenheit reading is 77"?
58. Given the situation of Exercise 57, let C denote the number of degrees
Centigrade and F the number of degrees Fahrenheit. Obtain an equa
tion that relates Q and F by substituting appropriate numbers for ¥%, F*,
P J? ("* f"*
Cn Ca in the following equation: r ~ ' * =?; tt •
128 Distance and Coordinate Systems Chapter 3
59. Starting with the equation obtained in Exercise 58, derive an equation
in simplified form that expresses F in terms of C.
60. Starting with the equation obtained in Exorcise 58 } derive an equation
in simplified form that expresses C in terms of F.
■ In Exercises 6165, a line with points and coordinates as marked in Figure
334 is given. Using the Two Coordinates Systems Theorem, x' may be ex
pressed in terms of x, first with A as Xi and B as X&* and then with B as Xj
and A as Xa The results are
* + 1 _ *  2 «* *>  r_5
« i^f=T^ w
52 w 16 25
A B X
4 f * » 1
2 ■> x
1 6 x'
Figure 3KM
If A = A, then x = 2, r' =  I, and Equation (a) beeomes
 y . , —  ~ n which reduces to = 0.
o + 1 5 — 2
61. Simplify Equation (a) if X = &
62, Simplify Equation (b) if X' = A.
63. Simplify Equation (b) if X = B.
64, Tlie sum of the left members of Equations {a) and (b) h
tf + 1 ar^6
6+1 16
which simplifies to
* + i . g'fl _ «* + i x'e _* + i  (x  6) 7 .
7 + 7 _ 7 f~" 7 = 7 = L
Add the right members of Equations (a) and (b) and simplify.
65. If you multiply the left member of Equation (a) by — 1 and add 1 to the
product, the result is x + * • (1) + L This simplifes to
y +J . ^j , T _ *  1 7
7 1 +  " 7 + 7
_ x + 6 x'  6 _ x /  6
7 : ~7 : 16'
which is the left member of Equation (b). Multiply the right member of
Equation (a) by — 1 and add 1 to the product. Show that the result sim
plifies to the right member of Equation (b).
3.7 Point* of Division 129
3.7 POINTS OF DIVISION
In I his section we use the Two Coordinate Svs terns Theorem b i help
us find the coordinates of the points on a given segment which divide
it into a given number of congruent parts or divide it in some other
specified way.
Definition 3.S The midpoint of a segment AB is the point
P on AB such that
AP = PB = \AB.
The midpoint of a segment is said to bisect the segment or to
divide it into two congruent parts.
Definition 3.7 The trivection points of a segment AS are
the two points F and Q on AB such that
AP = PQ = QB =
The trisection points of a segment are said to divide the seg
ment into three congruent parts. Similarly, points C. D, and
E on "KB such that
AC=CD = DE=EB = ±AB
are said to divide AB i nto four congruent parts. This idea may
be extended to any number of congruent parts.
A segment AT? is a set of points. You might think of a "path" from
A to B if you were to draw a picture of the segment. Although you may
think of AB as a path, be careful to remember that the segment from
A to B is the same as the segment from B to A, Indeed, AB = BA,
Sometimes, however, we want to consider B to A as different from
A to B. It might be helpful to think of trips and to consider the trip
from A to B as different from the trip from B to A. The point which is
onethird of the way from A to B is different from the point which is
onethird of the way from B to A. This leads us to the idea of a directed
segment. We think of a directed segment as a segment with one end
point designated as the starting point. A directed segment is known
as soon as the segment is known and the starting point is known. Our
formal definition follows.
130 Dfst«nc« and Coordfnate Systemi
Chapter 3
Definition 3.8 The directed segment from A to B, denoted
by AR is the set {Al, A}.
It is important to note the difference between the symbol for a rav
the symbol for a directed segment The ray symbol, as in AB a has
a complete arrowhead, whereas the directed segment symbol, as in
AB, has a half arrowhead. Directed segments are related to vectors,
and half arrows are frequently used in vector notation, Vectors are verv
useful in many branches of higher mathematics. Note that whereas
AB = M t it is not true that AB  BA. Note that
whereas
= (AB,A) = {BA i A} t
= {/U?, B} = {HA.BJ.
Definition 3.9 Let a directed segment AB and two points
P and Q on AB be given. If P £ AE, Q g AB y and
AP AO
PB ~ OB * ^ shown in ^ i § ure 3 "35, tnen ** and (J are said
to divide AB in the same ratio, P dividing it internally and
called an internal point of division, Q dividing it externally
and called an external point of division. The ratio ^ is the
PB
ratio of division.
A
Figure 335
f
B
Example 1 Given two points A and Bona line I with coordinates 4
and 20, respectively., as indicated in Figure 336, find the coordinate
of the point P on AB if P divides AB into two congruent parts.
20
<g>
► I
Figure 336
3.7 Points of Division 131
Solution: Let x be the coordinate of the desired point P. Then
4 < * < 20, x  4 s 20  x, 2x = 24, and s = 12,
Alternate Solution: Set up two coordinate systems as indicated in
Figure 337.
P B kf
# • — H
4 * 20 (S>
Figure $3f
The i .
4 _ fe
4 " 10
and ar = 4+ 16A: for every P on i*. Then F divides A5 into two con
gruent parts if k = ^ and
x = 4 + 16 • ^ = 4 + 8 = 12.
Example 2 Given two points A and B o n a line I with coordinates — 3
and 21, respectively, find the coordinates of the four points which di
vide it into five congruent parts.
Solution: Let S and $' be two coordinate systems on J with coordi
nates of several points as marked in Figure 338.
A P B
3 ! 3! 55 — M
* i ($'»
Figure 33S
Then
x (3) _ x+ 3 __ fc0
21 _ (_3) :: 24 1 0
= Jt
and x — 24* — 3 for every point P on /, The required points are the
points with ^coordinates \> » , y or. in decimal form, 0.2, 0.4, 0.6 t 0.8.
Now we compute the a: coordinates of the division points using the
equation x = 24k — 3. They arc
24(0.2)  3 = 4.8  3 = L8. 24(06)  3 = 144  3 = 11.4.
24(0.4)  3 = 9.6  3 = 6.6. 24(0.8)  3 = 19.2  3 = 16.2.
132 Distant* ami Coordtnite Systems Chapter 3
Example 3 Given two points A and B on a line I. with coordinates 3
and 21, respectively, find the coordinates of the points P and Q on
AB which divide AB internally and externally, respectively, in the
ratio Z.
Solution: Let $ and g' be two coordinate systems on I with coordi
nates of several points as marked in Figure 339.
Q A p b
T7. * * * ■ — H
{&) ** 3 * 21
Figure 349 ($') ** 0*1
Then, as in Example 2,
^J = k and x s 24k + 3
— 24
AQ 7
for every point F on J. Since ^ =  f A{> is less than QB and ^ is a
point on opp AB. The coordinates x and a:' of F and Q are computed as
follows:
AF=k
AQ =  ^
PB=1 k
p£ = 1  if
AF 7 k
A() 7 f
pB 8 "' 1tf
FB 8 1  k
r Ik s Sfc
7  7Jf = 8Jf
15* = 7
fifs 7
la
**= 24(^7) + 3 =
171
,= _24.X + 3 =
8.2
Check (using rcoordinatesK
AF 3(
PB 8.2 
8.2)
(21)
1L2 112 7
" 12.8 " 1.28 " 8
AQ 171
Qff 171  {
3
21)
K>S 42 7
192 ~ 48 ~ = 8
Example 4 Given two points A and B on a line I with coordinates 3
and 21, respectively, find the coordinates of the points R and Ton
AB which divide BA internally and externally, respectively, in the ratio
J. (Compare with Example 3.)
3.7 Writs of Division 133
Solution: Let $ and $' be the two coordinate systems on I with co
ordinates of several points as marked in Figure 340* Then, as in the
preceding examples, *  +" ^ = k and x — 24fc — 21 for ever}' point
HT 7 >
R on /. Since ~f = ^, BT is less than 7"A and T is a point on opp BA.
/A 8
< i
R
B
T
4 r
i
X
it
21
Figure 340
Then
BR = k
BT =  V
RA = 1  k
TA = 1  K
BR __ 7 _ fc
BI 7 jtf
flA "" 8 I 
k
TA" S 1  Jf
k4
15
(Why?)
£ = 7 (Why?)
x = 9.8
(Why?)
s? = 189 (Why?)
Check (using
xcoordinates).
Bfl _ 9,8 + 21 _ 11.2 _ 112 _ 7
RA 3 + 9.8 12.8 128 ~ 8
BT _ 21 H 189 ^ 168 = 1
TA 3+189
EXERCISES 3.7
In Exercises 15, A and B are points on a line I with given coordinates a and
&, In each exercise, find the coordinate of the midpoint of ~KB.
L o  5 S b = 27 4, a = S, b = 27
2, a = 5, h = 27 $. a = 0*b= 4.8
3. a = 5, h = 27
In Exercises 6 JO, A and B arc points on u line / with given coordinates a
and b. In each exercise, find the coordinates of the points which divide AB
into the given number, n, of congruent parts.
6. a = 3, b   7, n = 2 9. a = 8, b = 8, ft = 8
7. o = 3, b = 0, « = 5 10. a = 0, b = 10, n = 4
8. a = ], h = 79, n = 3
134 Distance and Coordinate Systems Chapter 3
■ In Exercises 1115, A and B are points with given coordinates a and b. In
each exercise, find the coordinates of the points P and Q which divide A~B
internally and externally in the given ratio r.
1L a = 10, b = 20, r = I 14. a = f, b = $, r = ^
12. a = 20, fr = 10, r = ? 15. a = , fc = J. r = ^
13. a = 26, fc = 0, f = f
16. Prove the following theorem.
THEOREM If tine coordinates of A and B are « and h t then the coordi
nate of the midpoint of AH is a + h .
17. Prove the following theorem.
THEOREM If the coordinates of A and B are n and h, then the coor
dinates of the bisection points of AB are ^ + b and  + ^ . .
3 3
18. A and B are points on a line / with coordinates and 1, respectively.
Find the coordinate of the point P which divides AB externally in the
ratio $&,
19. A and B are points on a line / with coordinates and 1, respectively.
Find the coordinate of the point Q which divides AB external] v in the
ratio tf$L,
20. A and B are distinct points on a line /. Is there a point P on AB*, but not
on AB, snch that P is the same distance from A as it is from B? That is,
is there a point P on AB which divides AB externally in the ratio y?
■ In Exercises 2 1 26, A and B are points on a line ( with given coordinates a
and b, respectively, P is the point which divides AB internally in the given
ratio r. la each exercise, select the statement from the righthand column
that is true.
21. a = 0, b = 1, r = ] (A) P is between the midpoint of A~E
and #
22. a = 0, b = 10, r ■ { (R) AP = 1
23. a = 10, b = 209, r = $$ (C) P is a bisection point of AB
24. a e 10, b = 211, r e jgj (D) BP = 1
25. a =  1.3, fc = 0,ra^ (E) P is the midpoint of OT
26. « = 0, h = 13, r = J£ (F) P is hetween the midpoint of ^
and A
Review Exercises 135
■ In Exercises 2730, A and B are points on a line I with coordinates a and h,
respectively. P is the point which divides A externally in the given ratio r.
In each exercise, select the statement from the righthand column that is
true.
27. a = Q.b = 1, r = ffifa (A) FA8 and BP = 1000
2S. a = (Kb=z hr= jggi (B) ARP and HP = 0,001
S8L Os^sUa U ^ L (C) ABP and BP = 1000
30. a = 0, h = 1, r = ^ (D) PAtf and AP = 0.001
CHAPTER SUMMARY
In this chapter we have developed the concept of DISTANCE between
two points and the concept of a COORDINATE SYSTEM on a line. We
mtroduced five postulates: DISTANCE EXISTENCE POSTULATE, DIS
TANCE BETWEENNESS POSTULATE, TRIANGLE INEQUALITY
POSTULATE, DISTANCE RATIO POSTULATE, and RULER POSTU
LATE, This chapter contains many definitions and theorems. A key defini
tion is the definition of a coordinate system. The climax of the chapter is the
TWO COORDINATE SYSTEMS THEOREM. In Section 3.7 we applied
the tools of this chapter to find points of division which divide a segment
internally and externally in a given ratio.
REVIEW EXERCISES
In Exercises 15, name the property that justifies the given statement.
13+4=4+3
2* (AB + CD) + KF = AH + (CD + EF)
3. If a — b> then 7? = a.
4. (.50 + 5) • 3 = 50 ■• 3 + 5 • 3
5. (5 + 3)7 = (3 + 5)7
In Exercises 611, name the postulate which justifies the given statement.
0, If P, Q, R are distinct collmear points with PQR, then PQ + QR =
PR.
7. If X and Y are distinct points and AH is any segment, then the distance
between X and Y in the distance function based on A2J is a positive
number.
8. If M, W> K arc three noncolliuear points, then RK + KW > RW.
136 Distance and Coordinate Systems Chapter 3
9. If A, B, C, D, E, F. G t H arc eight distinct points* then
AB (in ~EF units) _ AB (in C?r7 nnfts)_
CD (in EF units) CD (in C77 units) '
10. If A, ii, C, D t E, F, G, Ware eight distinct points, then
AB (in AT units) CD (in Eb units)
AB (in C77 units) " CD (in CT units)*
11. If A, B, C t D are four distinct points, then there is a unique coordinate
system on A B relative to CD such that the origin is B and the unit point
is A.
■ In Exercises 1220, A, B, and X are points on a line h a, h, % are their re
spective coordinates in one system j a\ h\ xf are their respective coordinates
in another system. In each exercise, (a) draw and label an appropriate figure,
(b) express x' in terms of x and simplify, and (c) express x in terms of x 1 and
simplify.
12. a a 0, b = 1, tf = 5, V = 8
13. a = 0, b = 1, d = 8, V = 5
14. c= 1, b = 0. a' = 5, tf = 8
15. a = 17, b = 16. a' = 5, &' = 8
16. o = 7, b = 3, ^ = 0, 6' = 1
17. a = 0, fc = 1, a' = 0, &' = 2
18. a = 0, b = 1, </ = a V = 1
19. a = 0, fo = 1, a' s 1, ¥ =
20. a = 0, /j = 100, «' = 1, fo' =
■ In Exercises 2130 t A, B, C f D. E, are points on a line I with coordinates
0, 1, 2. 3, 4, respectively, in a coordinate system on I In each exercise sim
plify the given expression. Each expression names a number.
21. AB (in AB units) = [T]
22. AB (in AT unite) = \f]
23. AC (in ATJ units) = [TJ
24. AD (in Al units) = (TJ
25. AD (in AC unite) s= [?]
26. AD (in A~E unite) s [?]
27. AE (in AB unite) = {T\
28. AB (in A~E unite) =
29. Ag(inXEui U ts)_
AD (in AT? unite)
30 AB (in IC unite) _ —
" AD (in AT unite) U
Review ExerciiB* 137
In Exercises 3140, A$ is a directed segment, P divides AB internally in the
positive ratio  , Q divides A~fi externally in the ratio  , and a and h are the
coordinates of A and B, respectively, in a coordinate system cmAB. In each
exercise, draw un appropriate figure und find p and q, the coordinates of P
and Q t respectively.
31. o = 0,ft = l,f=8,ia 1 36. a = 10. fe = 5.^ = 4
32.a = 0=l,r:=I p *=2 37. a = G t fc = 9,  = i
i' o
33. a = 0, h = 1, r = 2, a = 3 38. a = 6, b = 9,  = 3
34. a = 0. h = 1, r = 10, s = 5 39. a . = 1, b = 2, r = 5, * = 6
35. a = 10, h = 5, J a 4" 40. a = 1.3, & ss 7J* a A
s 4 * oo
In Exercises 4 147, X is a point on a line / and x and k are its coordinates in
two coordinate systems on I. The given equation tells how x and k are related
for every X on L In each exercise, copy and complete the given statement.
41. x = 3fe + 1 If k  0, then x =
42. x = 3*+i If fc= 1.thenx = 7]
43. x = 3fc +■ I < Jt < 1 if and only if fj] (a condition on x)
44.x=3Jt+l If k = 3, then a = [T
45. x m 3Jt +1 If Jt = 5. then x = [?]
46. a: = — 3fc f 1 k < — 3 if and only if JT] (a condition on x)
47. i = — 3fc + 1 fc > — 3 if and only if \T\ (a condition on i)
Fred Ward /Black Star
Angles,
RayCoordinates,
and Polygons
4.1 INTRODUCTION
In Chapter 3 we developed definitions and postulates for the con
cept of distance and for coordinate system* on a line. Bctweenness for
points is related to betweenness for red numbers through the idea of
a coordinate system on a line. The length of a segment is related to the
coordinates of its endpoints.
In this chapter we develop postulates and definitions for angular
measure or, as we usually call it, angle measure, and for raycoordinate
systems in a plane. Betweenness for rays is related to bctweenness of
the coordinates matched with the rays. The measure of an angle is re
lated to the coordinates of the rays that form the angle. Raycoordi
nates are useful in developing the properties of angles.
Chapter 4 concludes with a discussion of polygons and dihedral
angles. The idea of a polygon is a natural extension of the idea of a tri
angle, and the idea of a dihedral angle grows naturally from the idea
of an angle.
140 Angles, RayCoordinates, and Polygons
Chapter 4
4.2 ANGLE MEASURE AND CONGRUENCE
sen a pie is cut into four quarters of equal size as indicated in
Rgure 41, the rim is also cut into quarters. This is true regardless of
the size of the pie. If a pie is cut in the usual manner, then with each
piece that is less than half a pie there is an associated angle, sometimes
called the associated central angle. This angle is the union of the two
mys which have the point at the center of the pie as their common end
point and which contain the segments that are the cuts forming the
piece of pie. For a quarter pie w r e might think of the size of this angle
Figure 41
This corresponds to thinking of a revolution as the unit of meas
ure. If we adopted this unit, then the measures of the angles in our ge
ometry would be real numbers between and j. We prefer to think of
one revolution as equivalent to 360 degrees. Then the measures of
angles in our formal geometry will be numbers between and 180 as
suggested in Figure 42.
Figure 42
It is said that the Babylonians originated the system of measure
ment that is based on what we now call the degree as the unit. To them
the stars (except the sun) appeared to be fixed on a celestial sphere
that rotated about an axis once each day. The sun appeared to com
4.2 Angle Measure and Congruence 141
plete a circular path among the stars once each year (four successive
seasons}. They apparently knew that the length of the year was approx
imately 365 (lays, but, perhaps for convenience* took 360 days as their
"calendar" year. Considering that the sun traveled over a circular path
once each 360 days it was natural to divide that path into 360 equal
parts and consider each part as corresponding to one day and 90 parts
as corresponding to one season. In the early days of the Christian era,
the Greek mathematicians of the School of Alexandria divided the
circle into 360 equal parts and called each part a moira* This Greek
word was translated into the Latin word degradus, meaning *'a grade
or step from/' From this we get our word degree, meaning the first
step down from a complete revolution, or g£$ of a revolution,
Of course* we could use other units of angular measure such as rev
olutions or right angles. Some units that you may not have heard of are
mils, grads, and radians. There is no particular reason for using degrees
for angle measures other than the fact that this is commonly done and
lias been done for a long time. To make our development simpler, we
base our formal geometry of angle measure on just one unit of measure,
the degree.
POSTULATE 20 (Angle Measure Existence Postulate) There
exists a correspondence which associates with every angle in space a
unique real number between and 180.
Definition 4.1 The number which corresponds to an angle
as in the Angle Measure Existence Postulate is called the
measure of the angle.
Notation. The measure of A ABC is denoted by m L ABC.
Note that if the number of degree units in the measure of I. ABC
is 40, then ml ABC = 40 and not mLABC = 40*. If an angle is
marked 40° in a figure, it means that the measure of the angle is 40,
In Chapter 3 we agreed to call two segments congruent if they have
the same length. We make a similar agreement for angles.
Definition 4.2 Two angles (whether distinct or not) are
congruent angles, and each is said to be congruent to the
other if they have the same measure.
Notation. Z ABC s? ADEF denotes that I ABC and IDEF are
congruent.
142 Angles, RayCoordinates, and Polygons
Chapter 4
Although congruence and equality as applied to angles may seem
alike, they arc in reality different ideas. It is true lhat if LA = LB,
then LA ^ LB r but the converse is not true. Many pairs of congruent
angles are not pairs of equal angles. Remember that an angle is a set
of points and that two sets of points are not equal unless they consist of
the same point*;. For the angles suggested in Figure 43 we have
LABC = L CBA = LARK = L GBC = L GBK.
Figure 43
Also, LABC^ LDEF. Rut it is quite possible that mLARC =
mLDEE If this is true, then LABC m L DEE
The most common device for measuring, angles in informal geom
etry is a semicircular protractor with degree marks from to 180 evenly
spaced on the semicircular edge. To measure an angle such as in Figure
44a we can either place the protractor as indicated in (b) and read
the measure 20 directly or vvc can place it as indicated in (c) and ob
tain the same measure by subtracting 65 from 85.
Pbpus I 4
Another type of protractor is a circular one. (See Figure 45.) This
380degree protractor has advantages in drawing certain figures. In
using a 360degree protractor, as in using a semicircular one, it is pos
sible to obtain the measure of an angle from readings on its scale. This
and other properties of the protractor suggest the concept of a ray
coordinatc system which we define later.
4,2 Angle Measure and Congruence 143
Figure 45
EXERCISES 4.2
1. Copy and complete the following definition of congruence of angles.
If Z A = L B, then [fj and if »?£ A = mZB, then TJ.
2. (a) Use your protractor to construct three angles whose degree meas
ures are ©0, 90, and 135, respectively,
(b) Why is it not possible, using our definition of angle measure, to con
struct an angle whose degree measure is 240?
3. Use your protractor to find the degree measure of the angles shown.
144 Angfas, Ray Coordinates., and Polygon*
Chapter 4
4. In Exercise 3, did you find two angj.es thai arc congruent? If so, name
them and tell why they are congruent.
5. If Sandy Moser measures an angle and finds its. degree measure Lo be
60 and Bob Blake measures the same angle with the same protractor
and finds its degree measure to be 70, what can you conclude? What
postulate are you using as a basis for your conclusion?
Exercises 610 refer to the angles in Figure 46,
4
iur
E
U
1KT
rss
R
B
Figure 4*0
6. Using the notation of the figure, write two different names for equal
angles, that is, for the same angle.
7. Name two angles that arc congruent but not equal.
8. Write two names of angles such that the angles named are equal (and
therefore congruent).
9. Name two angles that are not congruent.
10, Can you name two angles that are equal but not congruent?
11. Using a protractor, measure the angles of the two triangles in the fol
lowing figure. List those pairs of angles that appear to he congruent
4.2 Angle Measure and Congruence 245
12. In the figure, the readings for certain rays with endpoint V are shown
in a circular protractor. Name four pairs of congruent angles, that is,
name four pairs of angles such that the angles in each pair are congruent
to each other.
13. Without using a protractor draw six angles whose degree measures you
would estimate to be 30, 45, 60, 90, 120* and 150, respectively. Alter
you have drawn each angle, measure it with your protractor and see
how good your estimate was.
14. Draw a triangle ABC so that m Z A = 43 t mlB = 57, and m Z C = 80.
15. Can you draw a triangle DEF so that mlD = 54, m£E = 67, and
mZF = 70?
16. Draw a triangle DEF so that mi. D = 54 and m/.E = 67. Use your
protractor to find m L F,
In Exercises 17 and 18, complete the proof of the following theorem.
THEOREM Congruence for angles is reflexive, symmetric, and
transitive.
Proof; Let Z A be any angle; then in Z A is a number and mAA = m Z A
by the reflexive property of equality. But if mLA = m/_A t then
LA^ L A by the definition of congruence for angles. Therefore congru
ence for angles is reflexive.
17. Prove that congruence for angles is symmetric
18. Prove thai congruence for angles is transitive.
1». (a) If LAm LB and LC^ IB, what can you conclude?
(b) What properties justify your conclusion in (a)?
146 Angles, RayCoordinates, and Polygons Chapter 4
4.3 BETWEENNESS TOR RAYS
Recall that if A, B, C are three distinct points on a line, then B is
between A and C if and only if
AB+ BC = AC
Also, B is between A and C if and only if B is between C and A. Be
tweenness for points is related to betweenness for "numbers througji
the definition of a coordinate system on a line and the Ruler Postdate.
For three distinct coplanar rays VA t VB, VC (with a common end
point) we want to develop a concept of betweenness based on our in
tuitive notions of symmetry, our experiences with protractors, and our
desire for additivity of angle measures in certain situations. Specifically,
if V£J is between V'A and VC, then we want VB to be between VC and
VA, Also, we want VB to be between VA and VC if and only if
m
Z AVB r m£BVC= m I AVC
These ideas suggest the following definition and postulate. Refer
to Figure 47 us you read them.
ix , jfrf , J&l
A and B are on tha / B and C are on the V A and C are on
same aide of VC / aanw aide of VA opposite sadea of VB
m
V r i i» betwaan V~X and V#
Figure 47
— ♦ — ► — > — >
Definition 4.3 If VA, VB, VC arc rays, then VB is between
VA and VC if and only if
1. A and B are in the same h airplane with edge VC.
2. B and € are in the same half plane with edge VA.
3. A and C are in opposite half planes with edge VB.
POSTLXATE 21 (Angle Measure Addition Postulate) If VA,
VB, VC are distinct coplanar rays, then VB is between VA and VC if
and only if
mLAVC = mZAVB + tnZBVC
4.3 Betweenness for Rays
We consider two important matters relating to Definition 4.3.
(A) Suppose that VJ? is between VA unci VC. Is VB between VC
and V??
(B) Suppose that V$ is between vX and V&, and that A', B', C
are any points, except V,.on VA, VB, VC, respectively, as in
Figure 48. Is vS' between vA' and VC??
In view of Definition 4,3, the questions raised in (A) and (B) amount
to the following, expressed in terms of Figure 48. Suppose
(1) A and B arc in the same
147
Figure 4S
half plane with edg<& VC,
(2) B and C are in the same
half plane with edge H,
(3) A and C are in opposite
half planes with edge VB.
Does it follow that
(4} C and B are in the same halfplane with edge VA,
(5) B and A are in the same halfplane with edge VC,
(0) C and A are in opposite halfplancs with edge VB?
Does it follow that
(7) A' and B' are in the same halfplane with edge VC,
(8) & and C are in the same halfplane with edge VA\
(9) A' and C are in opposite halfplanes with edge VB'?
Since (1) implies (5) and (7), (2) implies (4) and (8), and (3) implies
(6) and (9), it follows that the answer to questions (A) and (B) is Yes.
Thus betweenness for rays is symmetric just as betweenness for
points is symmetric. Indeed. VB is between VA and VC if and only if
VB is between VC and VA, and Q is between P and R if and only if Q
is between H and P. Also, betweenness for rays depends on rays, not
on the particular choice of points used in designating the rays.
Postulate 21 is consistent with these properties of betweenness.
Thus, referring to Figure 48.
nUAVC = mlAVB j mlBVC
mlCVA = mACVB + mlBVA
and
m£AVC = m£AVB + mLBVC
mLA'VC = mlA'VB' + m£B'VC
if and only if
(Why?)
if and only if
(Why?)
148 Angles, RayCoordinates, and Polygon*
Chapter 4
Example Consider the six coplanar and concurrent rays formed by
the three intersecting lines in Figure 49. Then study the following
two instances of betweenness in this figure and the four instances of
"not betweenness,"
Betweenness
1. VA is between VFand VB since (a) B and F are on opposite sides
of VA, (b) A and F are on the same side of VB. and (c) B and A
4 — ►
are on the same side of VF, From Postulate 21 it follows that
ml FVA + ml AVB = mlFVB.
2. VA is between VB and VF. Indeed, the three statements to check
are the same three statements that we just checked to verify
that VA is between VFand VB, From Postulate 21 it follows that
mlBVA + mlAVF=mlBVF,
which should not be surprising in view of the equation
in L FVA + m L AVB = m I FVB and the commutative prop
erty of addition,
Not Betweenness
1. VA is not between VF and VC, Why? Let us check. Are F and C
on opposite sides of VA? Yes, Are A and F on the same side of
VC? No. Of course, one '"So" in checking the three require
ments is enough to establish "not betweenness."
An alternate method of checking betweenness in this instance in
volves using Postulate 21. According to this postulate, VA is between
VF and VC if and only if
mlFVC = mlFVA + mlAVC
4.3 Betweenness for Ray* 149
But this equation is a false statement. Why? It is false because the right
side of the equation is a number, whereas the left: side is not. Indeed,
there is no such angle as Z FVC and hence there is no number such as
ntlFVC, (See Section 4.2.)
2. VA is not between V? and VB. Why not?
3. VX is not between V7?and vd, Why not?
4. VA is not between VA and VB. Why not?
EXERCISES 4.3
L In the figure below at the left, explain why ST is not between BA and
EC
A
2. In the figure above at the right, explain why VCis not between VA and
~VB.
3, In the figure for Exercise 2, is VB between VA and VC ? Is VA between
V$and VC??
4. In the figure at the right, if ABC.
then is BD between BA and BC?
Why?
D
4 — *
A
G
Exercises 59 refer to Figure 410, Assume that no two of the angles of the
figure are congruent. In each exercise, name the missing angle.
5. m Z AEB + ml EEC = m Z \j]
a mlAEC + ml CRD = mZ[T
7. ml ABC  ml ABE = ml^}
8. ml BED — ml\?} = mlBEC
9. mZ[0  mlECD = mlECB
Flgur* 440
ISO Angles. RayCoordinates, and Pofyf on* Chapter 4
■ Exercises 1013 refer to Figure 4 11 . Name the missing angle or number.
Figure 411 A
10. mlADE + nuLEDC = mZ(7]
11. mlDAB  nUDAC = mZ0
12, DE + EB = JTJ
13, AC AE = \7]
14 Given mlAVB = 35
mlAVC= 115
find mZBVCif
(a) VB is between vA and vS, and
(b) VA is between V7? and V?.
15. If t in a plane, mlAVB = 70 and mlBVC = 44, find mLAVC. Is
there just one possible value for m £ AVC? Illustrate with a figure.
16. In the figures below. tnLABC m/.EFG = 70 and mlDBG =
ml HFC = 25. Prove thai LABD^ lEFll
17, In the figures below, mlDEF s mLHKM = 120 and m/_GEF
mlHKN = 35, Prove that IDEG s ANKM,
18. In the figures below, mlPQS = mlDBC = m and mlSQR
mlABD = 50. Prove that IPQR ~ ZABC.
4.3 Betwa«nn*u for Hays 151
Tn the figure below, mZl = mZ3 and mZ2 = mZi Prove that
<LDEF = £DGF,
). challenge pnoni jcm. Let VB be between VA and VC and let R<p be
between /i? and /IS as shown in die figure below. It appears that all
three of the following statements might be true: (a) Z CVB s Z SK^>,
(b) Z BVA at Z ^>HP, and :c) Z C VA a / SRP. Prove that if any two
of these three statements are true, then the third one is also true.
Hint: There are three things to prove.
(A) If(a)aml(b)»then(c)
(B) If (a) and (c), then (b),
(Q If (b> and (c), then (a).
Proof of {A): There is a number r such that
m Z CVB = m / SRQ = r, Why?
There is a number $ such that
m / BVA = m Z £ilP = s. Why?
Then
miCVA = r+i= m L SRP> Why?
Then
ZCVAssZSRP. Why?
Now write proofs of (B) and (C).
152 Angles, RayCoordinates, and Polygon*
Chapter 4
4.4 RAYCOORDINATES AND THE PROTRACTOR POSTULATE
In the same way that the Ruler Postulate provides us with a mathe
matical, or abstract, ruler for assigning coordinates to points, \vc want
a mathematical protractor for assigning raycoordinates to rays. Wc
first define what is meant by a raycoordinate system. Then we adopt
the Protractor Postulate which amounts to an agreement that ray
coordinate systems do exist and that they are unique if we pin them
down in certain ways. The definition is based on experiences with
protractors.
Definition 4A (See Figure 412.) Let Vbe a point in a plane
«♦ A ray coordinate system in a relative to V is a onetoone
correspondence between the set of all rays in a with endpoint
V and the set of all real numbers x such that < x < 360
with the following property: If numbers r and s correspond
to rays VJR and VS in a, respectively, and if r > s, then
mZRVS = r — *
m£RVS = 360 (rs)
VR and VS arc opposite rays
if rs<180
if rs>!80
if r  s = 180.
Frgure4J2
Definition 4.5 The number that corresponds to a ray in a
given raycoordinate system is called the raycoordinate of
that ray. The ray whose raycoordinate is zero is called the
zero ray of that system.
Notation. We use cd VX as an abbreviation for raycoordinate of Va.
4.4 Ray Coordinates 153
Example 1 Figure 413 suggests a raycoordinate system where the
raycoordinates of several rays are given. For this example we have:
mAAVR = 70  = 70
mlAVH' = 360  (250  0) = 110
mZPVR =9070 = 20
mZ iW = 250  90 = 160
mlFVA = 360  (270  0) = 90
VR = opp vR' since 250  70 s 180
VA and VA' are opposite rays since 180 — = 180
Figure 413
POSTULATE 22 (Protractor Postulate) If a is any plane and
VA and VR are noncolHnear rays in a, then
(1) there is a unique raycoordinate system § in « relative to Vin
which cd vX = and <?<J VB = roZ A VB and
(2) if X is any point on the Bside of VA, then cd VX (in g) =
roZAVX
Example 2 Figure 414 shows a line VA and two rays vS and V? with
jB and X on the same side of VA..
V A Figure 414
If ml_ AVE = 50. there is a unique raycoordinate system in which
cd vZ = and cd VB = 50, If mlAVX = 105 and X and B are on
the same side of VA, then cd VX = 105.
154 Angles, RayCoordinates, and Polygon* Chapter 4
We now proceed to prove several theorems using our postulates
aibout angles.
THEOREM 4 J If Z A VB is any angle in a plane a and if £ is the
raycoordinate system in a relative to V in which cd VA = and
cd VB = m Z A VB, then the raycoordinate of VX is
(1) if VX = VZ*
(2) 180 if f£ s= app vX.
(3) between and 180 if X is on the Bside of VA.
(4) between 180 and 360 if X is on the notBside of VA.
Proof: Figure 415 consists of four parts which correspond with those
of the theorem.
V A
CD
Figure 415
1. In a raycoordinate system there is only one number matched
with each ray. Since VX = VA and since is matched with VA.
then is the number matched with VA.
2. According to the definition of a raycoordinate system two rays
are opposite rays if and only if their ray coordinates differ by
180. Since cd VA = 0, since VA = opp VA, and since every
raycoordinate is either or a positive number less than 360,
it follows that cd V? = 180,
4— * >
3. Since X is on the Bside of VA, then cd VX = mLAVX. Since
m Z AVX is a number between and 180, it follows that cd VX
Is a number between and 180.
4. Let X' be a point such that VX and VX' are opposite rays. Since
X is on the notBside of VA, it follows that X' is on the Bside of
VA. Then cd VX' is a number between and 180 by part 3, and
since cd VX is a number between and 360 which differs from
cd VX' bv 180. it follows that cd VX is between 180 and 360.
4.4 RayCoordinates 155
The argument in part 4 expressed in symbols consists of the fol
lowing steps:
0<cd VX<360
< cd VX' < 360
< cd V?' < 180
cdVX cd VX' = 180
cd V$ = cd V?' 4 180
180 < cd V? < 360
In working with a protractor we know that if we are given any an
gle, say Z DEF, and a ray VA on the edge of a halfpkne 3C, then we
can draw a ray Vfi with B in 3C so that IAVB = Z DEF. Our postu
lates permit us to do this in. the abstract as the following theorem
suggests.
IHEOREM 4,2 (Angle Construction Theorem) If Z DEF h any
angle, if VA is any ray, if 5C is any halfplane with edge VA, then
there is one and only one halfline VR in 3C such that /.AVBst
I DEF.
Proof: (See Figure 416.) Let us suppose that I DEF, VA, and 3C are
given as in the statement of the theorem. Let a be the plane that eon
tains ;1C and let F be any point in 3(1, Let § he the unique raycoordi
nate system in o relative to V in which
cd VA = and cd VP = mlAVF.
E p Figure 416
Let m Z DEF = b. Then < b < 180 and there is exactly one ray VB
with B in 3C such that cd VB = b. Why? Then
ml AVB = b = ml DEF
and VB is the unique halfline in 3C such that LAVB^ Z DEF.
The following theorem relates betweenness for rays and between
ness for coordinates'.
156
Angles, RayCoordinates, and Polygons Chapter 4
THEOREM 4,3 If a raycoordinate system in which cd VA = 0,
cdVB = b,cdVC = c with c < 180 is given, then VB is between
VA and VC if and only if b is between and c.
Proof: Figure 4 1 7 stipes ts the "given' ' or hypothesis of our theorem ,
We prove two things,
L If b is between and c, then
VB is between vZ and v£
2. If VJ?is between vA and v3,
then h is between and c.
Proof of 1:
In;, ,iv t17
Statement
L e < 180
2. c >
3, < b< c <
4. mZAVB= 60
5, m/_BVC=cb
a mZAVC = c 
7, (b  0) + (c  b) = c 
8, ml AW + mZBVC =
mZAVC
9, VB is between VA and
VC,
IteofafSt
Statement
1. VB is between VA and vS.
2. VA, VB, VC are distinct
rays.
3. B is on the Csidc of VA
4. 0, b* c are distinct num
bers.
5. < c < 180
1. Hypothesis
2* All raycoordinates are non
negative numbers,
3. Hypothesis and steps 1, 2
4. Definition of raycoordinate
system
5. Definition of raycoordinate
system
6. Definition of raycoordinate
system
7. Properties of real numbers
8. Substitution (steps 4, 5, 6,
and 7)
9. Angle Measure Addition
Postulate
L Hypothesis
2, Definition of betweenness
for rays
3* Definition of betweenness
for rays
4. Definition of raycoordinate
system and step 2
5. Hypothesis
4.4 RayCoordtoitM 157
6. < h < 180
7. m£AVB = b0
mLAVC = c 
8, mZAVB + mLBVC
mlAVC
9, 6 + mZBVC = c
10, mZBVC>0
11, fc < c
12, 0<fc<c
6. Steps 3 f 4, 5 and definition
of a raycoordinate system
7. Definition of a raycoordi
nate system
8. Angle Measure Addition
Postulate
9. Substitution (steps 7 and 8)
10. Angle Measure Existence
Postulate
11. Steps 9 and 10
12. Steps 6 and 11
THEOREM 4.4 (Angle Measure Addition Theorem) If distinct
rays VB and VC are between rays VA and VD and if Z AVB s
ZCVD, then LAVC sa LBVD. '
Proof: Figure 4 J 8 suggests two
possibilities. We prove both cases
at the same time.
v a v a
F*w*4lS (.) (b)
Suppose that VB and VC are distinct rays as in Figure 4I8> From
the Protractor Postulate and Theorem 4.3 it follows that there is a
unique raycoordinate system § such that cd VA = 0, < cd VD <
180,0 < cd v3< cd VD, andO < cd VC < cd VB. Let cd v8 = b,
cdVC = ccdvB = d.
We must show that if A AVB m Z CVD t then Z AVC S Z BVD.
Since
mLAVB = h = fc
mlCVD= d c
m£AVC = c
m£BVD = db
our problem amounts to proving that if b = d — c._ then c =z. d — b.
Suppose then that b = d — c. Using the Addition Property' of Equal
ity, we may add c — h to both sides of this equation. The result is
which simplifies to c = d — h, the desired conclusion.
Notice tho similarity of Theorem 4.4 to Theorem 3.4.
158 Angles, RayCoordinates, and Polygon* Chapter 4
COEOLLAR 1 4,4, I If distinct rays VB and VCare between rays
VA and V/3 and if Z AVC as Z BVD, then I AVB m Z CVD,
Proof: If VB and V(? are between vX and VD, then vS and VB are
between VA and VD. The corollary follows immediately from The
orem 4.4 by interchanging VB and VC, that is, by renaming ray VB
as ray VC and renaming ray VC as ray VB.
COROLLARY 4,4,2 If VA, VB. VC. VD are distinct coplanar
rays such that AVD, B and C are on the same side of AD, and
Z AVB at L CVD, then Z AVC =s Z BVD.
Proof There is a unique raycoordinate system in which cd VA = 0,
cd VB = b, cd W= c, cd v3  180, b^ c, < h < 180, and
0<c< 180. Then
mlAVB = b0=b t
ml AVC = c = c,
mLBVD 1806,
mZCVD= 180 c.
Since Z AVB ss Z CVD, then b =s 180  c. It follows that b + c =
180, c = 180  fe, ml AVC = mlBVD, and ZAVC^= ABVD.
EXERCISES 4.4
Exercises 110 refer to u raycoordinate system in which the numbers as
signed to VA, V$. VC?, VD, V? ( and V?ure 0. 28, 47, 139, 263, and 319,
respectively. In Exercises 29, compute the angle measures using the given
raycoordinates.
1. Draw a figure to illustrate the given situation.
2. mlBVC 6, mlFVE
3. m/ BVF 7, mZAVE
4. mlCVE 8. ml BVD
5. mlDVE 9. mlCVF
10. What can you say about rays VD and PF?
11. Let three distinct concurrent rays VA t VB, VA' such that VA = opp VA'
be given. Prove that
ml AVB + mlHVA' = 180,
4.4 RayCoordinates 159
12. la the situation of Exercise 11, explain why it is incorrect in our formal
geometry to say that
mLAVB + mLBVA' = mLAVA'.
Kxercises 1319 refer to a raycoordinate system in which the numbers as
signed to VA, VB, VC, and VD are 0, b t c, and d, respectively, where
0<fr<e<ti< 180. In Kxercises 1419, express the angle measures
using the given raycoordinates,
13. Draw a figure to illustrate the given situation,
14. mLAVC 17. m/CVD
15. mLBVD 18. m/AVD
16. mLBVC 19. mlAVB
hi Exercises 2020, the same situation as in Exercises 1319 is given except
that < b < c < 90 and 270 < d < 360. In Exercises 2126, express the
angle measures using the given raycoordinates.
20. Draw a figure to illustrate the given situation.
21. mLAVC £4, m/CVD
22. mLBVD 25. mLAVD
23. mLBVC 26, mLAVB
Exercises 2738 refer to Figure 419. If the raycoordinates of vX, vS, V6 t
VD, VE, and VF are as .shown in the figure, name the theorem, definition,
or postulate that justifies the statement in the exercise.
U5G4
27. VD is between VS and VE. 33. LBVC == Z D VE
28. m Z CVE = ml CVD + 34. Z BVD m Z CVE
m/DVE
29. V? is between V?and VB. 35. mZ AVE = 140  = 140
30. mLBVF = ml BVC + 36. m I BVF = 180  40 = 140
m Z CVF
31. ml BVC = 70  40 = 30 37. LAVE BS Z BVF
32. mLDVE = 140  110 = 30 38. LAVB B LEW
160 Angles, Ray Coordfnales, and Polygons
■ Exercises 39 and 40 refer to Figure 420. Use Corollary 4.4.2.
Chapter 4
i— *
Figure 420 S
39. If ZM\W at / HVS, name an angle congruent to AMVR.
40. If /LSVNs* ZflVM> name an angle congruent to ZSVR.
41. In the figure below, L DBA S L CBE, What can you conclude about
Z 1 and Z2? What are you assuming from the figure about the points
A t B t C t 0, El About the points A, B, C in particular? About D and E
in relation to AC?
B D
iL
B
42. In the figure below, Z 1 — Z 2. What can you conclude about LBAE
and Z CAD? Name the theorem you are using. What assumption are
you making about the figure?
4.5 SOME PROPERTIES OF ANGLES
Two distinct intersecting lines form four angles as suggested by
Figure 421. Two angles such as a and c or h and d t which appear "op
posite" each other in the figure, are called vertical angles. Two angles
such as a and b or h and c are called a linear pair of angles. We state
these ideas more precisely in the following definitions.
Figure 421
4.5 Some Properties of Angles 161
Definition 4.8 Two angles are called vertical angles if and
only if their sides form two pairs of opposite rays.
Definition 4. 7 Two angles are called a linear pair of angles
if and only if they have one side in common and the other
sides are opposite rays.
Note that two angles are vertical angles if their union is the union
of two distinct intersecting lines. Note also that two angles are a linear
pair of angles if their union is the union of a line and a ray whose end
point lies on that line*
THEOREM 4. 5 Vertical angles are congruent.
Proof: Let AB and CD intersect at V to form two vertical angles
A AVC and Z BVD as in Figure 422. Let a be the plane determined
by A, V, and C Let % he the unique raycoordinate system in a relative
to V in which cd VA = and cd VC = mL AVC. (Which postulate
tells us there is one and only one raycoordinate system with these
properties?) For convenience, let cd VB = b, cd VC a c t cd VD = d.
b Figure 4SS
The rest of the proof follows from the definition of a raycoordinate
system.
b = 180 and d = c + 180 Why?
mlSVD =d b=(c+ 180)  180 = c = mlAVC
L AVC ^ L BVD
Definition 4,8 Two angles (distinct or not) are complemen
tary, and each is called a complement of the other if the sum
of their measures is 90. Two angles (distinct or not) are sup
plementary, and each is called a supplement of the other if
the sum of their measures is 180.
162 Angles, RayCoordinates, and Polygons Chapter 4
THEOREM 4.6 If two angles form a linear pair of angles, then
they are supplementary angles.
Proof: Let a linear pair of angles, Z AVB and Z BVC, in plane a as
indicated in Figure 421 be given. Let § be the unique raycoordinate
system relative to Vin a such that cd VA = and cd VB = m L AVB.
4 • if * ►
Figure 423 A V C
Then cd VJB < 180 and cd v2 = 180
ml AVB + mlBVC = (cd VB  0) + (180  cd VB) = 180.
Therefore Z. AVB and ZBVC arc supplementary.
Notice that Exercise 11 of Exercises 4.4 is Theorem 4.6.
THEOREM 4. 7 Complements of congruent angles are congruent .
Proof: Let Z A and £ B be two congruent angles, let ZC'bea com
plement of Z A, and let Z D be a complement of ZB. Then
mlA + niLC 90
m/_8 + mlD = 90
mZA := roZB
r»'ZA h mZC = mZB + mLD
mlC = mlD
IC^ ID
THEOREM 4.8 Supplements of congruent angles are congruent.
Proof: Assigned as an exercise,
A special ray associated with an angle is its midray. Here is a defini
nition for a midray.
Deftnitfon 4.9 A ray is a midray of an angle if it is between
the sides of the angle and forms with them two congruent
angles. A midray of an angle is said to bisect the angle; it is
sometimes called the bisector of the angle or, briefly, the
angle bisector.
4.5 Some Properties of Angles 1 63
Tt appcais from Figure 424 that an angle should have one and only
one midray, This suggests our next theorem.
Figure iM
THEOREM 4. 9 Every angl e has a unique midray.
Pmof: Let IAVC as in Figure 425 be given. Let § he the unique
raycoordinate system in which cd VA a and cd VC = c
= ml AVC. Then there is a unique ray V£? such that cd VB s b m ^
Why? 2
ml AVE + m Z BVC = (fc  0) 4 (<?  I)
= c = mZAVC
VB is between VA and VU.
mZAVB = fc 0= b
mlBVC = c~ b=2b  h = h
mlAVB = ml8VC
Therefore VB is tetween VA and VC, and VB forms with V& and VC
two congruent angles. Therefore VB is a midray of IAVC,
We prove next that I AVC has only one midray. Suppose that VD
is a midray of IAVC. Let cd VD = d,
Then VD is between VA and V?. Why?
< d < c by Theorem 4.3
mlAVD= d^0 = d
mlDVC= v  d
c — d = d, c = 2d, d= 4 = b
VD = VB Why?
Therefore the midray is unique.
164 Angles, Ray Coordinate*, and Polygons
Chapter 4
EXERCISES 4,3
L In the figure several rays and their raycoordinates are marked. Com
pute the measures of the following angles,
(a) LAVB
(b)
LAVC
(c)
LAVH
(d)
LAVG
(e)
IAVF
(f)
LDVC
(g)
LDVB
m
LUVA
9
IDVH
(1?
LDVC
:sn :i
2, In a raycoordinate system the numbers assigned to VA. VB, VC, VD
are 0, 30, 130, 180, respectively,
(a) Draw a figure to illustrate the given situation.
(b) Compute several angle measures and use them to prove that VB is
between VA and VC.
(c) Is VB between VA and VD? Justify your answer.
(d) IsVA between V/sand VC? J ustify your answer.
(e) Is VC" between VBand V0P Justify your answer.
3. In a raycoordinate system die numbers assigned to VA and VB arc
and 100, respectively. All of the rays and angles in this exercise are in
the. plane determined by the points A, V, and B.
(a) If m/.AVC = 50 and if B and C are on the same side of VA, find
cd VC.
(b) If m L AVC = 50 and if B and C are on opposite sides of VA, find
cdvS.
(c) If m Z BVD h 150 and if A and D are on the same side of VB, find
cdvS.
(d) If m / BVD b 150 and if A and D are on opposite sides of VB, find
cd V5.
(e) If cd VE = 200, us vS between VA and VE?
(f) If cd VE = 200, is VA between V# and VE?
4.5 Some Properties of Angles 165
(g) If cd vS = 200, is Vj£ between V/l and VA?
(h) If erf VF = 281, is VB between V# and V??
(i) If cd V? = 281, is VA* between vS and V??
(j) If cd VF = 281. is VF between VS and VA*?
£k) Rod cd VK if VK b the midray of Z AVB.
4« Given a raycoordinate system in which the numbers assigned to VA,
VN, VD, VY are a, b, c, d, respectively, let VA\ VN\ v5\ v9* t bo the
rays opposite to VA, VN, VD, VY, respectively. Assume that <£, a <
b<ic < cf <C 1 SO. Derive formulas in terms of ft, b, c. d for the follow
ing measures.
(a) mXAVN (e) mlDVA
(b) mlAVY (f) mZJDVY
(c) mLAVy (g) mLDVN
(d) mLAVU (h) mtDVN
5, Referring to the figure, describe in your own words the following sets.
(a) The union of VA and m (c) The union of Z AVB and Z BVC.
(b) The union of VA and VB. (d) The union of Z AVB and / RVC.
Referring to the figure, describe in your own words the following sets.
(a) The intersection of VR and VF. \^p Xt
(b) The intersection of ZiiVP and ZPVX
(c) The intersection of Z H VP and I XVY. *_
(d) The intersection of LTVX and I XVP.
7. Let AM be the edge of a halfplane £JC that contains points C and D. As
sume that A, C, D arc noncoilincar and that B, C, D are noneollinear.
Describe the following sets.
(a) The intersection of 3C and AB.
(b) The intersection of Z CAP and 3C.
(e) The intersection of Z CAD and AB.
(d) The intersectiOD of Z CAD and Z CBD.
(e) The union of 3C and /LB.
(f; The union of 3C and Z CAD,
(g) The union of 0C and ZAGB.
(h) The intersection of JC and ZACJ3.
166 Angles, RayCoordlnatM, and Potygorta Chapter 4
8. Let four coplanar rays VA, VB, VC", VD he given such that VB is be
tween VA and VC, VC is between VB and VD, and VD = opp vl as
indicated in the figure. Complete the following proof that
mLAVB + mLBVC + mlCVD 180.
Why?
Why?
Proof: m L BVD = w Z BVC + mL CVD
mLAVB + mL BVD = 180
Therefore [TJ.
9. Let four coplanar rays VA, VB, VC, VD be given such that vS is be
tween VA and vS, v8 = opp V&, and m£AV€ > mLAVB as indi
cated in the figure. Justify each step in the following proof that
mLAVB + ml BVC + mZCVD = ISO.
. e
B
4
B
• M80
v
Proof: Let § be the unique raycoordinate system in which cd VA =
cdvts mLAVB = b < 180.
*— * — >
Then B and C are on the same side of Ay, I ,et cd VC = c. Then
< c < 180, where a = mLA VC. Then < b < c < 180. Therefore
mZ AVB + m I BVC + mL CVD
= (b  0) + (o  b) + (180  c) = 180,
10. In the figure AB and c3 intersect at O
forming four angles, If tnLAOD = 133,
find
(a) ml CO A
(b) mlBiX:
(c) mZBOD
11. If LA and Z B are supplementary angles and i is a number such that
mZA = 3* + 6 and mZ J? = &r + 12, find the measures of Z A and
Z B. Check your results by finding the sum of these measures.
12. If LF and Z Q are complementary angles and y is a number such that
mLF = y + 30 and mL{) s (/ — 30, find the measures of LF and
Z (X Check your restilt by addition.
13. Twice the measure of an angle is 24 more than five times the measure of
its supplement. Find the measure of the angle and check your result.
4.5 Some Properties of Angles 167
14. Find the measure of an angle if its measure is twice the measure of its
complement,
15. Find the measure of an angle if its measure is onehalf the measure of its
complement.
16. If / A and Z B are both congruent and supplementary, find the mea
sure of each.
17. If Z A and Z B arc both congruent and complementary, find the mea
sure of each.
18. The figure shows tlirec eoplanar lines and six angles marked a„ b } c t d,
x, and y. Complete the following state
ments, (There are several correct re
sponses for some items.]
(a) aandi? are a [3 pair of angles,
(h) a and b arc angles.
(c) 6 and c are \J} angles.
(d) If a sj x, then b s* [T]
19. Given an angle L AVB, let 3C he the sot of all points that are on the
same side of VB as A. that is, JC is the Aside of VB. Let X be the Bside
of VA. Make a sketch showing JC by sliading with vertical h amines and
X by shading with horizontal halflines. How is the interior of £AVB
marked in your sketch? Is the interior of LAVB the intersection of 5C
and X. or the union of 3C and X, or some other set related to 3Q and X?
20. In the proof of Theorem 4.7, justify each of the five equations and the
congruence.
21. Prove Theorem 4.B.
22. In the figure below, AC is the midray of Z BAD and VS is the midray
of LRVT. If lBAD=z IRVT, prove that ZJ?AC= ZRV5.
£3. Speaking informally, does the result of Exercise 22 prove that "halves"
of congruent angles are congruent?
24. Using Theorem 4.8 and the reflexive property of congruence for angles,
write an alternate proof of Theorem 4.5.
25. In the figure at the right,
Z2=s Z3.
Prove that
Z1^Z3
Z4^ Z5.
168 Angles, RayCoordinates, and Polygon*
4.6 INTERIORS OF ANGLES
In Chapter 2 wc defined the interior of Z ABC as the intersection
of two halfplanes, the Cside of A B and the Aside of BC.
Another way to think of the interior of an angle is in terms of the
rays between the sides of the angle. From our definition of between
ness for rays in Section 4.3, we know that if VD is a ray between the
sides of LAVC y then every point of VD except V (that is, the hairline
VD) lies on the Asidc of VC and on the Cside of AV. Hence VD is con
tained in the interior of the angle. What we have shown for VD is true
for every ray between VA and VC. For convenience let (R denote the
union of the interiors of all rays between VA and VC and let $ denote
the interior of / AVC. We have shown that (Ft lies in 3 t that is, that
(Red.
In Lemma 4. 1 0. 1 we shall show that if a point is in §, it is also in (ft.
Then we shall have (R C d and £ C (R, and hence $ =s (R, which we
state formally in Theorem 4.10.
LEMMA 4.10.1 If a point is in the interior of an angle, then it is
an interior point of a ray between the sides of that angle.
Proof: Let D be a point in the interior of Z.AVC, and let a be the
plane containing the points A, V, C, D. Then it follows from the defini
tion of the interior of an an^le that (1) A and D arc on the same side of
VC and (2) D and C are on the same side of VA. We shall show diat VD
— ? >
is between VA and VC y and hence that D is an interior point of a ray
between the sides of L AVC,
Let § and §' be ray coordinate systems in a relative to V with co
ordinates as indicated in the table and such that <C c <C 180.
< a' < 180. Since < c < 180 and since C and D are on the same
side of VA, it follows that < d < 180. Since < d < 180 and since
A and D are on the same side of VC, it follows that < & < 180.
S 3L
cdVA a'
cdVD d d'
ud VC?
c
4.6 Interiors ot Angles 169
Suppose that c = d. Then d' = (Why?), D lies on VC t and D is
not in the interior of Z.AVC. Since this contradicts the hypothesis of
the theorem, it follows that c^d.
Then < d < c < 180 or < c < d < 180. Suppose that
< c < d < 180.
Then it follows from Theorem 4.3 that VC is between VA and VD and
from Definition 4.3 that A and D are on opposite sides of VC. However,
A and D are on the same side of VC, (See (1) in proof.) Since this con
tradicts the Plane Separation Postulate, il follows that
< d < c < 180.
Using Theorem 4*3 and Definition 4.3 again, we deduce that YD is
between VA and VC and hence that D is an interior point of a ray be
tween the sides of £ AVC,
We have proved the following theorem,
THEOREM 4.10 The interior of an angle is the union of the in
teriors of all rays between the sides of the angle.
Another way to consider the interior of an angle is in terms of the
segments whose endpoints he on the sides of the angle. (Sec figure
426.)
Figure 426
Given Z ABC, let D be any point on BA except B and let E be
any point on BC except B< Let P be any interior point of the segment
DE. In other words, P is any point between D and E. Since all of DE
except E lies on the Aside of BC and all of DE except D lies on the
Cside of AB t it follows that Plies in the interior of A ABC. Thus we
have the following theorem.
THEOREM 4.11 If AB is a segment joining an interior point of
one side of an angle to an interior point of the other side, then the
interior of AB is contained in the interior of the angle.
170 Angles, RayCoordinates, and Polygons
Chapter 4
4.7 ADJACENT ANGLES AND PERPENDICULARITY
A linear pair of angles is a special case of a pair of coplanar angles
having one side in common and interiors which do not intersect. (See
Figure 427.) It is convenient to introduce a special word for angles
with th is property.
4
Figure 427
Definition 4.10 Two coplanar angles are adjacent angles if
they have one side in common and the intersection of their
interiors is empty.
"fl
Figmc 42S
In Figure 428, L AVE and Z B VCare adjacent angles and Z AVE
and Z AVC are adjacent angles. Although Z DWE and Z EWF arc
adjacent angles, note that Z DWE and Z DWFaxe not adjacent angles.
All of you have a background of experience with right angles, per
pendicular lines, acute angles, and obtuse angles. Following are the
formal definitions for these terms.
Definition 4.11 An angle whose measure is 90 is a right
angle. An angle whose measure is less than 90 is an acute
angle. An angle whose measure is greater than 90 is an obtuse
angle, (See Figure 429.)
Sight EUjgle
Figure 42M
AouU nriKle
Oi'LusL' im^Uti
4.7 Perpendicularity 171
THEOREM 4. 12 If the two angles in a linear pair are congruent,
they are right angles.
Frew?/: (See Figure 430.) Let the measure of each angle in the linear
pair be r. It follows from Theorem 4.6 that r + r = ISO, Therefore
r = 90 and each of the angles is a right angle.
Figure 430
THEOREM 4.13 Any two right angles are congruent.
Proof; Every right angle has a measure of 90. Hence all right angles
have the same measure and hence all right angles are congruent to each
oilier.
Definition 4.12 If the union of two intersecting lines con
tains a right angle, then the lines are perriendicular.
If Z A VB is a right angle, then AV and VB are perpendicular
(See Figure 431.)
mcs.
• A
~V
A
B'
>A'
Figure 431 Figure 432
It b easy to show that two perpendicular lines form four right
angles. Let A A' and BB' be perpendicular lines which intersect at V
as indicated in Figure 432 and let L AVB be a right angle. Then
LA'VW is a right angle since LA'VB' and Z AVB arc vertical angles
and vertical angles are congruent. Also, L AVB r is a right angle since
I AVB' and A AVB form a linear pair. Then ml AVB = 90,
mLAVB + ml AVB' = 180, and therefore ml AVE' = 90, Why is
Z A' VB a right angle?
172 Angles, RayCoordinates, and Polygons
Chapter 4
When are two rays perpendicular? It seems natural enough to say
that two rays are perpendicular if the lines which contain them are per
pendicular. We extend this idea to any combination of segments, rays,
and lines in the following definition.
Definition 4. 13 Two sets, each of which is a segment, a ray,
or a line, and which determine two perpendicular lines are
called perpendicular sets, and each is said to be perpendicu
lar to the other.
Notation. We write © _L (B to mean that <2 and (B arc perpendicular
sets, Wc may read <$. _L (B as "'d is perpendicular to (&' f Note that if
a 1 (B, then OS 1 4
If d and (B are perpendicular sets, then & is contained in some line
I, (B is contained in some line m, and I JL m. Since perpendicular lines
are distinct intersecting lines, it Follows that if Q, _L (B. then d and (B
have at most one point in common. As indicated in Figure 433, the
intersection of two perpendicular sets is a set consisting of at most one
point.
A ray porpoodieular to a
line; intersection is empty
ray perpendicular tn n
ray; interaeclion \s hmi;!v
Figure 433
4.7 Perpendicul»rtty 173
In a figure two perpendicular sets may be marked with a little
"square comer" as illustrated in Figure 434.
3.
Figured
A typical exercise in informal geometry involves constructing a per
pendicular to a line at a point on it. Our formal geometry is sufficiently
developed now so that we can prove that such perpendiculars exist.
THEOREM 4. 14 For each point on a fine in a plane, there is one
and only one line which lies in the given plane, contains the given
point, and is perpendicular to the given line.
Proof: See Figure 435. Let P be a point on line I in plane a. Let 0C
he one of the two halfplanes in a with edge I. There exists a point A
different from P on /, Why?
B
X
• R
Figure 4io
Let R be any point in 3C. Let § be the unique raycoordinate
system in a relative to P in which cd PA = and cdPR = m/ APR.
Let PR be the unique ray with cd T3 = 90. LAPS is a right angle
andPB± PA. Why?
Therefore there is at least one line in a through P perpendicular to t.
Let m be any line in a through P and perpendicular to I Then
m U /contains four right angles. One of these right angles is the union
of FA and a ray, say PC, such that PC? C 3C.
m£APC = 90
0<cdpd< 180
cd PC = 90
PC = PB Why?
m = PE Why?
Hence there is only one line in a through P and perpendicular to L
174 Angles, RayCoordinates, and Polygon*
Chapter 4
In connection with Theorem 4.14 there is an interesting question
to consider. If / is a line and F is a point not on I, is there one and only
one line through P that is perpendicular to T? (See Figure 436.) Maybe
there is no line m such that m L i. Maybe there is Just one such line m.
Maybe there are several* We shall prove later that in our formal ge
ometry there is one and only one line m through P and perpendicular
to I. The point in which m intersects I is called the foot of this perpen
dicular from P to I.
p
+i
Figure 4*36
It is important to emphasize that all lines considered in Theorem
4.14 lie in one plane. Tf we remove this restriction, it seems reasonable
that there are many lines perpendicular to a given line at a given point
on it, as suggested in Figure 437. This idea comes up again in a later
chapter.
Figure +a?
EXERCISES 4,7
Let a raycoordinate system § in a plane n relative tea point Vbe such that
the raycoordinates of VA, VB, VCarc a, b> c, respectively In Exercises 15,
given the values of a, 6, G, determine whether or not B is in the interior of
IAVC.
I a = t b = 90 t c = 175
2. a = 0, b a 90, c = 185
3. a = 0,b = 270, c = 185
4. a = 0, h  270, c = 175
5. a = 90, h = 260, c =
4.7 Perpendicularity 175
In Exercises 616, A t B, Care three noncollinear points and D is an interior
point of BC.
6. Is D an element of the interior of Z J9AC?
7. Is D an element of the interior of Z BCA?
8. Is D an element of the interior of / CAB?
9. Is J? an element of the interior of Z CAU?
10. Is C an element of the interior of Z DAB?
11. Is AB a subset of I ABC?
12. Is AB a subset of I ABC?
1 3. Is the interior of A~fi a subset of Z ABC ?
14. Is SUa subset of /.ABC?
15. Is AD a subset of die union of Z.BAC and its interior?
16. Answer each question under "reason" with a definition, a postulate, or a
theorem in the following proof that mZ DAC + mLDAB = m£BAC.
atcment
(a) D is an interior point of BC.
(b) D is in the interior of
ABAC.
(c) AD is between A~B and AC.
(d) mLDAC + mLDAB =
mLBAC
(a) Given
(b) Why?
(e) Why?
(d) Why?
■ In Kxereises 1 722, state which of the following descriptions, (a) to (e), cor
rectly identifies the perpendicular sets in the given diagrams.
(a) Two perpendicular rays,
(b) Two perpendicular segments,
(c) A line and a ray perpendicular to each other.
(d) A ray and a segment perpendicular to each other,
(e) A line and a segment perpendicular to each other.
17. <
+
20.
21.
22.
t
\
176 Angles, RayCoordi nates, and Polygon* Chapter 4
■ In Exercises 2327, supply the missing word so that each sentence is Inie.
23. If the angles in a linear pair of angles are congruent to each other, I hen
each of them is a \T\ angle.
24. If the measure of an angle is less than 90. then it is an \T\ angle,
25. If the measure of an angle is greater than 90. then it is an [T angle.
20. If one angle of a linear pair of angles is acute, then the other one is [TJ.
27 If A ABC is a right angle, then BA and S?are [?] lines.
■ Let a set of coplanar rays be given as in Figure 438 such that m Z VAX =
130, m/MVR = 65, m£SVM = 30, VT is the midray of ZflV?>, VS is
between VM and VR t VR is between VS and VD. Use this information to
find the measure of the given angle in Exercises 2832.
2& IRVS
29. ZDVR
30. ZDVT
31. ZDVS
32. ZrVTVf
Figure 43S
■ In Figure 4*39, A3, CD f EF are coplanar lines that intersect at V and
mLAVC = 32.3 and m£AVF = 151.7. Use this information to find the
measure of the given angle in Exercises 3337.
33. ZCVF
34. LFVB
35. LBVE
36. AEVA
37. AEVC
Figure 4.30
■ In Exercises 3842, Z AVR in plane a and a raycoordinate system in « such
that cd VA = 270 arc given. In each exercise, supply the two numbers that
are as close together as possible and that make the resulting statement true,
38. If A AVB is obtuse and cd vS < 90, then cd VB is a number between
(Taiid{¥J.
39. If Z AVB is acute and cd VS < 270, then cd VB is a number between
{?] and [T].
40. If Z AV.fi is obtuse and cd vB > 90, then cd VB is a number between
\T\ and [7.
41. If I AVB is acute and cd v3 > 270 } then cd V$ is a number betwoen
[T]and[7].
42. If/ AVB is a right angle, then cd V$ is [7] or [
i
4.8 Polygons 177
43. Copy and complete the proof of the following theorem. Draw a figure
to help you understand the result.
THEOREM U I A VB, Z BVC, and L CVA are three coplanar angles
such that the interiors of no two of them intersect, then the sum of the
measures of these three angles is 360.
Proof: In plane A VB there is a unique raycoordinate system § relative
to Vin which cdvX = and cd VB = ml AVB. From the [T] Postulate
it follows that m Z AVB is less than \J] t hence that cd VB is less than [T.
By JTj the interiors of A. AVB and /.AVC do not intersect. Therefore
C and B lie on [T] (the same side, opposite sides) of AV. Therefore the
raycoordinate of VC, call it c> is [T] than 180. Since the interiors of
Z BVC and Z AVC do not intersect, it follows that c is less than h + [r\.
Then
mLAVB= h 0 = b
mlBVC = \2\
mLCVA = fT]
mZAVB + mlBVC + mLCVA = b + (c  b) + = 30Q
44. In the proof of Theorem 4J4 there is a statement that m J ! is the
union of four rigjit angles. One of these angles is described in terms of
PA arid a ray PC such that PC C 3C. Describe the other three angles
in a similar way,
4.8 POLYGONS
As stated before, triangles are three sided polygons and quadri
laterals are foursided polygons. A polygon is a plane figure with n ver
tices and n sides, where n is an integer greater than or equal to 3, The
figures shown in Figure 440 are polygons, but the figures shown in
Figure 441 are not.
Ffpir»44J
178 Angles, RayCoordinates, and Pelyfont
Our formal definition is as follows,
Chapter 4
Definition 4.14 Let n be any integer greater than or equal
to 3, Let Pi, P 2 . . . . , P n u P* he n distinct coplanar points
such that the n segments P,P 2l P^, P^iA, Jyi have
the following properties:
1. No two of these segments intersect except at their
endpoints.
2. Xo two of these segments with a common endpoint arc
collinear.
Then the union of these n segments is a polygon. Each of the
n given points is a vertex of the polygon. Each of the n seg
ments is a side of the polygon.
Tf n — 3, the definition of a polygon yields a triangle; if n — 4, it
yields a quadrilateral. Sometimes a polygon with n sides is called an
ngon. For example, a polygon with 13 sides is a 13gon. The following
list gives the names commonly used for the polygons having the num
ber of sides indicated. You should learn these names.
• umber of Sides
N
ante
10
12
'triangle
Quadrilateral
Pentagon
Hexagon
Heptagon
Octagon
Decagon
Dodecagon
Notation. The polygon whose vertices are Fi, F& . . . a P„ and whose
sides are TJ^ t Pp^, .... P« ^.P^is called the polygon PjP 2 . . . P„.
Definition 4. 15 Two vertices of a polygon that are end
points of the same side are called consecutive vertices. Two
sides of a polygon that have a common endpoint arc called
consecutive sides. A diagonal of a polygon is a segment
whose endpoint s are vertices, but not consecutive vertices,
of the polygon.
4.8 Polygons 179
Most of our work with polygons is restricted to polygons like those
ill Figure 442,
Figure 44£
We usually exclude polygons like those in Figure 443 because they
are not convex polygons.
Figure 443
The polygons that interest us are the convex polygons, The follow
ing is our formal definition.
Definition 4. 16 A polygon is a convex polygon if and only
if each of its sides lies on the edge of a half plane which con
tains all of the polygon except that one side.
You in ust be careful not to confuse the idea of a convex polygon
with that of a convex set as defined in Section 2.5. A convex polygon
is not a convex set of points, although the union of a convex polygon
and its interior is (sec Definition 4,17).
Figure 444 illustrates Definition 416, It shows the halfpkncs as
sociated with the sides as required in the definition.
180 Angles, RayCoordinates, and Polygon* Chapter 4
Figure 445 shows a quadrilateral ABCD that is not convex. Neither
the Dside of BC nor the Aside of BC contains all points of the quadri
lateral not on BC. Draw a figure to convince yourself thai every triangle
is a convex polygon.
Ffgurc+*5
Definition 4.17 The interior of a convex polygon is the in
tersection of all of the Imlfplanes, each of which has a side of
the polygon on its edge and each of which contains all of the
polygon except that side.
Figure 446 illustrates this definition for a convex quadrilateral
ABCD.
3C = Cside of A~§
6 = Dside of BC?
g = Aside of cB
DC = Bside of Ha
Interior of ABCD = 3C (1 £ O $ PI 3C
Definition 4, 18 An angle determined by two consecutive
sides of a convex polygon is called an angle of the polygon.
Two angles of a polygon are called consecutive angles of the
polygon if their vertices are consecutive vertices of the
polygon.
For a triangle, any two of its vertices are consecutive vertices, any
two of its sides are consecutive sides, and any two of its angles are con
secutive angles. For a quadrilateral we use the word opposite when
consecutive is not applicable as in the following definition.
Definition 4.19 If two sides (or vertices, or angles) of a
quadrilateral arc not consecutive sides (or vertices, or angles),
then they are opposite sides {or vertices, or angles) and each
is said to be opposite the other.
4.9 Dihedral Angles 181
Figure 447 illustrates this definition. For this example we have
A and C are opposite vertices.
B and D arc opposite vertices,
AB and CO are opposite sides,
BC and 73^ are opposite sides,
LA and / C are opposite angles,
L B and L D are opposite angles.
Figur* 447
4.9 DIHEDRAL ANGLES
Angles are plane figures. Every angle is a subset of a plane. Closely
related to the idea of an angle is the idea of a dihedral angle. Sometimes
we say plane angle when we want to emphasize that an angle is not a
dihedral angle. A plane angle is the union of two noneollinear rays hav
ing the same endpoint. A dihedral angle is formed by two halfplanes
and a line. Here is the formal definition.
Definition 4*20 If two noncoplanar halfplanes have the
same edge, then the union of these halfplanes and the line
which is their common edge is a dihedral angle, The union of
this common edge and either one of these two halfplanes is a
Face of the dihedral angle. The common edge is the edge of
the dihedral angle.
A dihedral angle is suggested by Figure 448. In this diagram,
B and C are poinfe on the edge, A is a point in one face but not on the
edge, D is a point in the other face but not on the edge. A suitable
name For this dihedral angle is ABCD or LABC~T>. The letters
at both ends of this symbol are names of two points, one in each face
but not on the edge; the two letters in the middle are names of distinct
points on the edge.
Am
Figure 448
182 Angles, RayCoordinates, and Polygon*
Chaptor 4
The same way two intersecting lines form four angles, two inter
secting planes form four dihedral angles as indicated in Figure 449.
LAVE
LDVC
LCVD
Z.DVA
Figure 449
Definition 4,21 Two dihedral angles, such as AUVB and
DIJVC in Figure 449, which have a common edge ami
whose union is the union of the two intersecting planes are
vertical dihedral angles.
EXERCISES 4.9
I, Copy and complete the following definition of a pentagon:
Let F, Q, R t S, T be five df&ttoct eopbmar points such that the five seg
ments [T, EH, EH* S [D have the following properties.
(a)[lJ
mm
Then the [T] of these five segments is a pentagon.
In Exercises 27, one of the sides of hexagon ABCDEF is given. In each
exercise, name two sides which arc consecutive with the given side.
5. DE
2. AB
3. BC
4. CD
6. IF
7. FA
4.9 Dihedral Angles 183
8. Copy and complete the following proof that if ABCD is a convex quad
rilateral, then the interior of AC is contained in the interior of the
quadrilateral.
Proof; Let if denote t he interior of AU a nd let $ denote the i nterior of
the quadrilateral.
U
C Cside of AB
C Csidc of AD
C Aside of BC
C Aside of CD*
But g is the intersection of [Tj. Therefore $ gj.
9. (a) Draw a convex pentagon and all of its diagonals,
(b) How many diagonals does a convex pentagon haver
(c) Does the interior of each diagonal of a convex pentagon lie in the
interior of the pentagon?
10. Draw a pentagon which is not convex. Label it and explain why it is
not convex.
11. Draw all of the diagonals of the pentagon you drew in Exercise 10
I low many diagonals are there?
Figure 450 shows a labeled cube. In Exercises 1216, a quadrilateral whose
vertices arc vertices of this cube is given. la each exercise, write a suitable
name, using four of the letters from the figure, for a dihedral angle contain
ing the given quadrilateral.
12. ABCD
13. ABFE
14. BCGF
15, CGHD
16. I'Mlll
B Plgvn 450
Exercises 1721 also refer to Figure 450. Base your answers on examining
the figure. In some cases the formal geometry that we have developed up
to this point is inadequate to prove the correctness of these answers.
17. Is ABFH a dihedral angle?
18. Is ABFC a dihedral angle?
19. Is ABFE a dihedral angle?
20. Is ABFD a dihedral angle?
2L Is GHEC a dihedral angle?
184 Angles, RayCoordinales, and Polygon*
Chapter 4
In Exercises 22 and 23, a dihedral angle ABCD* a line I in plane ABC
through A and not intersecting BC, and a line m in plane DBC through D
and not intersecting BC arc given.
22, Draw a figure to illustrate this situation.
23. Explain why I and m do not intersect.
24* Figure 451 shows two planes, a and & intersecting in line AB. How
many dihedral an^es arc formed by these two intersecting planes?
Name them.
25. I>efine the interior and exterior
of a dihedral angle.
26. Define adjacent dihedral angles.
27. Name a pair of adjacent dihe
dral angles in Figure 451.
28. Name a pair of vertical dihedral
angles in Figure 451 .
Figure 451
CHAPTER SUMMARY
The central theme of this chapter is properties of angles. We intro
duced three new postulates; The ANGLE MEASURE EXISTENCE POS
TULATE, the ANGLE MEASURE ADDITION POSTULATE, and the
PROTRACTOR POSTULATE. The definitions include the following.
CONGRUENT ANGLES
MEASURE OF AN ANGLE
BETWEENNESS FOR RAYS
RAYCOORDINATE SYSTEM
VERTICAL ANGLES
LINEAR PAIR OF ANGLES
RIGHT ANGLE
OBTUSE ANGLE
ACUTE ANGUS
COMPLEMENTARY ANCLES
SUPPLEMENTARY ANGLES
MIDRAY OF AN ANGLE
ANGLE BISECTOR
ADJACENT ANGLES
PERPENDICULAR SEGMENTS,
RAYS, AND LINES
POLYGON
CONVEX POLYGON
DIHEDRAL ANGLE
Review Exarclses 185
This list of terms should remind you of many of the ideas and theorems
of tliis chapter. Raycoordinates have been used extensively in developing
the ideas and in proving the theorems that are included Theorems 42
(ANCLE CONSTRUCTION THEOREM), 4,4 (ANGLE MEASURE AD
DITION THEOREM). 4.5, 4.6, 4.7, and 4.8 are important for the work that
you will do in Chapter 5. Be sure that you know die statements of these
theorems.
REVIEW EXERCISES
In Exercises 110, write definitions for the following terms.
1. Congruent angles
2. Vertical angles
3. Adjacent angles
4. linear pair of angles
5. Complementary angles
6. Supplementary angles
7. Right angle
8. Acute angle
9. Obtuse angle
10. Dihedral angle
In Exercises 1 1 20. a, b, c are the raycoordinates of VA, VB, V T C, respec
tively. In each exercise, determine if one of the three rays is between the
other two. If one is, name it. If none is, write "none,* 1
11. a = 0, b = 10, c = 170
12. a = 0, b = 10, c = 180
13. a = t b= 10, c = 190
14. a = 0, b = 10, c  200
15. o = 350, h= 50, c= 110
16. a = 270. b = 180, c = 135
17. a = 270. b m a c = 135
IS. a = 180, h = 90, a = 20
19. a = 135, b = 315, c =
20. a. = % b m 300, c = 100
186 Angles, RayCoordinates, and Polygon! Chapter 4
■ In Exercises 2130, a and b sire the raycoordinates of VA and VB, respec
tively. In each exercise, compute the measure of / AVB.
2L a = 38, b = 106
22. a = 300. fo = 150
23. a = 300, b = 100
24. a = 359, h = 1
23. o ^ 270, h = 100
26. a = 38, & = 50
27. a = 198 s b =
28. a = 15, fc = 300
29. a = 6, b = 40
30. a = 315, b = 345
■ In Exercises 3140.
cdVA = cdvf= 150
tfd W? a 10 cd VK  200
afVC?=20 cdv£ = 2S0
ar\S = 30 crfVM=300
crfv£=40 «/V>f=305
cd V?=50 cdVP = 310
erf v2 = 75 erf VQ = 315
cd VH = 90 erf V7* = 325
erf V?= 100 crfVS = 330
In each exercise, determine whether or not the given statement is true.
31. LBVffm IJVL
32. Z7VH = LDVE
33. Z SVB = Z UVF
:I4, Z B VE and Z DV7J are supplementary angles.
35. Z CVE and Z SV7 are supplementary angles.
36. Z.KVL and LAVE are complementary angles.
37. W? is the midray of L DVJ.
38. V0 is the midray of L MVP
39. V^ k the midray of Z KVE,
40. Z SVil is an acute angle,
41. Explain how betweenness for rays is related to betvveenness for ray
coordinates.
Review Extremes 187
42. Explain bow betweenncss for rays is related to the addition of angle
measures.
43. Ii LA and Z B are complementary angles and if mZA = X and
mlB= 3* + 10, findr.
44. If Z T and Z W are supplementary angles and if mLT = X and
mZW=3*+ 10, find x.
45. Use raycoordinates to prove that vertical angles are congruent.
46. Use raycoordinates to prove that the angles of a linear ]>air are
supplementary,
47. If AVA' r BVff, cdVA = 143, cd vS<cdVA, and tX' I «"#\
find ed VB, cd W, cd VB'.
4 S. 1 f A BCD is a convex quadrilateral and £ is a point not in the plane of tiie
quadrilateral, which of the fallowing are names of dihedral angles:
ABCB, ABCE, DBCE, EADC?
49. Explain why a convex polygon is not a convex set.
50. Explain why a dihedral angle is not a convex set.
Chapter
CtHirtvsy of Quebec Government TouriH Office
Congruence
of Triangles
5.1 INTRODUCTION
Suppose you placed a sheet of carbon paper between two sheets of
paper and drew a picture of a triangle on the top sheet Will the carbon
copy of the triangle that appears on the second sheet be the same size
and shape as the one you have drawn on the top sheet? The idea of two
physical objects being carbon copies of one another is what we have
in mind when we say that the two objects have the same size and shape.
Obviously, we cannot take a physical object such as a house and, using
carbon paper, make a carlxm copy of it We can, however, take the
same set of blueprints that were used in constructing one house and
construct another house that is exactly like it, diat is, having the same
size and shape as the first one.
In this chapter we are concerned witii the "size and shape" of ge
ometrical objects such as segments, angles, and triangles. These geo
metrical objects are not the physical objects that we draw on our paper.
They are the mathematical objects which exist in our minds and whose
properties have been described in our postulates, definitions, and the
orems. The mathematical concept corresponding to "same size and
shape" is congruence. We, have already defined congruence of seg
ments in Chapter 3 and congruence of angjes in Chapter 4. Since all
segments have the same shape, we say that two segments are congruent
190 Congruence of Triangle*
Chapter 5
if they have the same size, that is, if the measures of their lengths are
equal. Similarly, two angles have the same size and shape or are con
gruent if their measures are equal
5.2 CONGRUENCE OF TRIANGLES
In order to arrive at a definition of congruence for geometrical tri
angles, let us consider again two physical triangles, one of them a car
bon copy of the other. Consider first a triangle that has sides of three
different lengths, a scalene triangle. The definition we will obtain ap
plies, however, to all triangles regardless of their shapes. Figure 51
shows two scalene triangles, A ABC and its carbon copy. AA'B'C
If you arc to make A ABC "fit" on AA'B'C, you must match up the
vertices of the two triangles according to the scheme in Figure 51:
A <— * A',
B «— ► B',
C <— + C.
This matching is called a onetoone correspondence between the two
sets of vertices. The correspondence between the vertices of the two
triangles also gives a correspondence between the sides of the triangle.
A75
ISC
A'W,
If a correspondence between the vertices of two triangles is such
that the corresponding angles and corresponding sides of the two tri
angles are congruent then the correspondence is called a congruence
between the two triangles. The correspondence we have been discuss
ing is a congruence between A ABC and AA'&C. On the other hand,
5.2 Congruence of Triangles 191
it is possible to write a correspondence between the vertices of the two
triangles that is not a congruence. For example, the correspondence,
A « — * B\
B +* C t
C ^^ A\
is not a congruence between the two triangles shown in Figure 51 be
cause, by this matching of vertices, it is not possible to make A ABC
coincide with &A'WC. Write four more correspondences between
the vertices of A ABC and AA'FC such that no two are the same and
such that none is a congruence between the two triangles.
It is convenient to write a correspondence, such as
A < — ► A\
B <— » W,
C<— » C,
on one Une as ABC ± — * A'B'C When this notation is used, it is un
derstood that the first letter on the left corresponds to the first letter
on the right of the double arrow, the second corresponds to the second,
and the third corresponds to the third as shown below.
tf
BC <— ► A'B'C
V .. »
1
Note that there arc several ways of writing this same correspond
ence, For example, both BAC 4 — * B'A'C and CBA « — » CB'A'
name the same correspondence as ABC * — * A' WO.
The foregoing discussion about corresponding vertices of physical
triangles can be made to apply equally as well to abstract geometrical
triangles. Thus, if ABC i — > DEF is a correspondence between the
vertices of any two geometric triangles, this correspondence provides
us with three pairs of points. We are interested primarily in these points
as vertices of the angles of the triangles and as endpoints of the sides of
the triangles. In connection with this correspondence we speak of six
pairs of corresponding parts. Three of these six pairs are pairs of angles:
LA *^ ID,
L B < — > L E,
IC ^— £R
The other three pairs are pairs of sides:
AB < — > Z5E,
BC < — ► Iff,
au ^ ra
192 Congruence of Triangles Chapter 5
We are now ready lo state the definition of a congruence between
two triangles.
Definition 5.1 Two triangles (not necessarily distinct) are
congruent if and only if there exists a onetoone correspond
ence between their vertices in which the corresponding parts
are congruent. Such a onetoone correspondence between
the vertices of two congruent triangles is called a congruence.
If ABC < — > DEF is a congruence between A ABC and ADEF,
then we write A ABC at A DEF and note that the following six state
ments are true:
I A aZD, AB^DE,
in? lE r BC=zEF,
ACstLF, AC = DF.
We can also say that, for A ABC and A DEF, if all of these six state
ments are true, then ABC * — * DEF is a congruence, or what means
the same thing, AABC as ADEF, In view of the definition of con
gruent triangles, we sometimes say that "corresponding parts of con
gruent triangles arc congruent,"
If AABC == ADEF, explain why each of the following six equa
tions is true:
m£A = m£D t AB = DE,
m£B = m£E, BC = EF,
tn£C=mlF t AC = DF.
Note that if AABC and ADEF arc distinct triangles and if
ABC < — > DEF is a congruence, then it is correct to write AABC ss
ADEF t but that it is incorrect to write AABC = A DEE The state
ment "AABC ==; A DEF" is a short way of saying "ABC « — * DEF
is a congruence;" it is a statement about a onetoone correspondence
between the vertices of two triangles. The statement "AABC =
ADEF" is a statement that two sets arc equal; it means that AABC
and ADEF are names for the same triangle.
Note also that
if AABC ss A DEF t then ABAC == A EDF,
For AABC B ADEF means that ABC * — > DEF is a congruence;
ABACS A EDF means that BAC < — * EDF is a congruence; and
ABC 4 — > DEF and BAC < — > EDF are two ways of describing the
same onetoone correspondence between vertices. On the other hand,
if ABC i — > DEFh a congruence but ABC < — > FED is not a con
5.2 Congruence of Triangles 193
gruence between the two triangles, then it is incorrect to write
AABCm A FED.
In drawing figures it is convenient to label congruent angles and
congruent sides of triangles with the same number of marks as shown
in Figure 52.
c
Figure 52
of the six congruences indicated in the figure are
LA^LF, IB^IE, /.C^ID.
Name the three pairs of congruent sides indicated in the figure. The six
congruences tell us ABC < — > FED and A ABC m AFED.
It is also helpful to label the side that Is opposite a given angle in a
triangle with the same number of mark* used in labeling the angle, as
has been done in Figure 52. For example, in AABC t L C and side^fB
(which is opposite LC) both have the same number of marks.
EXERCISES 5.2
Exercises 16 refer to Figure 53.
D Figure W
EOF is not a, congruence.
A V3 * * VI
1. Is AABC ^ ADEF? Why?
2. Explain why the correspondence ABC * — ►
3. Is it correct to write AABC ss AEDF?
4. Is it correct to write ABC A = AEFD? Why?
5. Write four more Statements of congruence between the two triangles,
each of which foikiws immediately from the fact that ABC * — » DEF
is a congruence.
6. Write three correspondences between the vertices of the two triangles
such that each is not a congruence,
194 Congruence of Triangles
Chapter 5
7. The following figure shows two scalene triangles with corresponding
congruent sides and angles marked alike.
Copy and complete the following correspondences so that the re
sulting statements are true,
LKP * — * \T\ is a congruence,
LPK « — > \J} is a congruence,
KLP < — * [T] is a congruence.
KPL 4 — » [TJ is a congruence.
PKL * — > is a congruence,
PLK < — * [f] is a congruence.
& In Exercise 7, why is the correspondence LKP * — * KST not a
congruence?
9. In the figure. AA£Ca APQB. Copy and complete the following
statements by supplying the missing symbols.
The correspondence A[T * — * \]}QR is a congruence.
Lh*k LP AE^^l
10. In the figure, AABCs A DBG. List the
six pairs of corresponding, congruent parts
of these two triangles.
5,2 Congruenco of Triangles 195
1 1 . In Exercise 10, if £ ABC and Z DBC are distinct coplanar acute angles,
does my B? bisect I ABD? Wiry?
12. In Exercise 10 (with the figure appropriately modified), if L ABC and
L DBC are coplanar obtuse angles, does ray BC bisect Z AHDP Why?
13. The figure below shows eight triangles. If two triangles look congruent,
assume that they really arc congruent. Write congruences between such
congruent pairs of triangles.
14. Without using a figure, list the six pairs of corresponding, congruent
parts for the triangle congruence
AEFK =s AABT.
15. Without using a figure, list the six pair* of corresponding, congruent
parts for the triangle congruence
ARPSss ALSR
196
Congruence of Triangles Chapter 5
16. For A ABC and A CAS, it is true that AH  $C, AC.  SA, #C as CA,
Z A as IS, Z C as Z A, and Z A as Z G. Write a statement of con
gruence between these two triangles.
17. If &DEFis a scalene triangle* prove that the statement
ADEFsz ADFE
is false,
18. If A AST as ASIA, what special property does A A ST have? Draw a
suitable figure for A AST".
19. If A/Jfitf as A MEN, what special property does ALMA r have? Draw
a suitable figure for ALMN,
In Exercises 20 and 21, complete the proof of the following theorem.
THEOREM Congruence for triangles is reflexive, symmetric, and
transitive.
Proof: Let A ABC be any triangle. Then IA as ZA» Z B £* Z A t and
ZCas ZC by the reflexive property of congruence for angles. Also
AA=?AB, WlrsBCt and AC as AC
by the reflexive property of congruence for segments. Therefore A ABC as
AAAC by the definition of congruent triangles. Therefore congruence for
triangles is reflexive.
20. Prove that congruence for triangles is symmetric
21. Prove that congruence for triangles is transitive.
22. In the figure at the right,
CD 1 AB,
AC as BC,
AD = BD t
LACD as Z BCD,
m£A = SO,
mZB= 00.
Prove that A ACD as A BCD,
23. In the figure at the right,
Af is the midpoint of §7,
Mi s the midpoint of 7ZF.,
KS as EJ>
mZS=_r»ZJ,
W J KE f
IE ± KE.
Prove that A SKM as A JEM.
5.3 "IfThen" Statements 197
5.3 "IF THEN" STATEMENTS AND THEJR CONVERSES
Many of our definitions, postulates, and theorems have been stated
in the "ifthen" form. They have been statements of the type "If p,
then tf," where p and q are statements, (Remember that a statement is
a sentence which is either true or false.) In other instances we have
used the phrase "if and only if ** in the statement of some of our defi
nitions, postulates, and theorems.
As tin example of the use of the phrase "if and only if consider die
definition of congruent segments given in Chapter 3: Two segments
are congruent if and only if they have the same length. This statement
is a conjunction of the following two statements:
L Two segments are congruent if they have the same length.
2. TVo segments arc congruent only if they have the same length.
Statements 1 and 2 can be restated in the "ifthen 1 * form as follows:
3. It two segments have the same length, then they are congruent.
4. If two segments are congruent, then they have the same length.
We note two important things about these last two statements.
(a) Each is written in the "ifthen" form. The ifclause of each
statement is the thenclause of the other. (This also means that
the thenclause of each is the ifclause of the other.)
(b) Both statements may be used in proofs. For example, if we
know AB as n? and want to establish AB = CD, we can use
statement 4 to justify our writing AB = CD, On the other hand,
if we know AB = CD and wish to establish AB s CD t we can
use statement 3 for justification.
The two statements, J and 2, or alternatively 3 and 4, are called
converses of each other. That is, the statement "If p, then q" is called
die converse of the statement "If q, then p" and "If q, then p" is the
converse of "If p.. then q" The converse of a statement in the "ifthen"
form can be obtained by interchanging the if and thenclauses.
If p r then (j.
If g, then p.
For example, the converse of the statement:
"If 1 live in Seattle, then I live in the state of Washington"
is the statement
"If I live in the state of Washington, then I live in Seattle."
198 Congruence of Triangles Chapter 5
It is evident from this example that a statement and its converse need
not both be true.
When a definition is given in the "ifthen" form, it is understood
mat the statements of the definition and its converse are both true. As
an example, consider Definition l.l of Chapter 1:
Space is the set of all points.
This definition can be restated in the "ifthen" form as follows:
1. If S is space, then S is the set of all points.
The converse of (1) is
2, If S is the set of all points, then S is space.
Thus the complete definition is: If S is space, then S is the set of all
points; and if S is the set of all points, then S is space. The "if and only
if' form (which is logically equivalent to the complete definition) is;
S is space if and only if S is the set of all points.
Although a definition in * "ifthen" form always implies the converse
statement, this is certainly not true of all postulates and theorems. We
discuss the "ifthen" form of a theorem more fully in Section 5.4.
As an example of a postulate whose converse is not true, consider
Postulate 2 (The LinePoint Postulate).
"Every line is a set of points and contains at least two points."
We can restate this postulate in the "ifthen" form as follows:
"If I is a line, then / is a set of points and contains at least two
points."
The converse of this statement is
"If I is a set of points and contains at least two points, then t is a
line."
Clearly, this last statement is false. There are many sets of points such
as planes that contain at least two points and that arc not lines.
Consider the following theorem proved in Chapter 4.
"Vertical angles are congruent."
In the "ifthen" form, this theorem can be stated as follows:
"If LA and L B are vertical angles, then LA === LB"
Surely the converse of this theorem, staled as follows, is false:
"If LA=* LB, then LA and / B are vertical angles."
5.3 "IfTtwn" Statements 199
We conclude this section with some remarks about definitions and
truth. A definition in our formal geometry is accepted as a true state
ment. Why is "Space is the set of all points" a true statement? It is true
"by definition." Definitions help us communicate. It is helpful to have
one word that means the same thing as "the set of all points/* It is help
ful to have one word to describe several points that all He on the same
line. Why do eollinear points all lie on the same line? They do, by
definition.
EXERCISES 5.3
In Fxerdses 15, a definition iu "ifthen" form is given. Write its converse.
L If the points of a set are eollinear. then there is a line which contains all
of them.
2. If there is a plane which contains all the points of a set, then those
points are coplanar,
3. If point A is between points B and C, then rays A3 and A C are opposite
rays.
4. If an angle is a right angle, then its measure is 90.
5. If W is the midray of LABC. then BP is between BA and §C and
ZAflFSB IFBC '
In Exercises 68, write the definition of the given phrase using the "if and
only if form.
6. Acute angle
7. Linear pair
8. Vertical angles
In Exercises 916, is the statement true or false? Write the converse of the
statement. Is the converse true or false?
9. If two sets are convex, then their intersection is convex.
10. If two angles are right angles, then they are congruent.
11. If two angles are complements of congruent angles, then they are
congruent,
12. If 5» is the interior of an angle,, then S is a convex set.
13. If two angles are congruent, then they are supplements of congruent
angles.
14. If I! st CD, then WstAB.
15. If AABC^ AKLM, then AKLM St A ABC,
lft If AABCss ADEF and AD£Fs ARST, then A ABC as ARST.
200 Congruence of Triangtos Chapter 5
5.4 THE USE OF CONDITIONAL STATEMENTS IN PROOFS
As you have seen, many of the theorems have the form "tip, then
</," where p and q are statements. Not all theorems are stated in this
way, however, because it is sometimes easier to state them otherwise,
but every theorem can be rostated in the "If p, then q" form.
A statement of the form "If p> then tf is called a conditional The
ifclause (the p statement) Is called the hypothesis and the thenclause
(the q statement) is called the conclusion. In order to understand math
ematical proof better, we examine how such statements are used in
proofs
Let us consider the following statements concerning two numbers
x and y about which we know nothing except what we are told in the
statement (A).
(A) lix= y, then x 2 = y*.
We see that (A) involves two statements.
(B) x = y (the hypothesis)
(C) x 2 = y 2 (the conclusion)
Even though we do not know the replacements for x and y, can we say
anything about the truth of statements (A), (B) s and (C)? Do we know
that ( B) is true? Do we know that (G) is true? What about statement
(A)? Your experience in working with numbers should help you to see
that even though (B) need not be true and that (C) neeA not be true,
(A) is true. Thus a conditional may be true even though its hypothesis
and conclusion are not. Replace x and y with several pairs of numbers,
some of which are equal and some of which are not. You should find
that in those cases where you chose unequal numbers for x and y t both
(B) and (C) are false and in 'those cases where you chose equal numbers
for x and y, both (B) and (C) are true. But, in every case, (A) is true.
Let us examine one such case where the replacements for x and y
are unequal numbers. Suppose we replace x with 2 and y with 1 , Then
statement (B) becomes 2 = 1 and statement (G) becomes 2 2 = I 2 , or
4 = 1. Of course, both statements (B) and (C) are false. However, if
we accept the hypothesis that 2 = 1, use the multiplication property
of equality to write 3 • 2 = 3 • 1, or 6 = 3, and use the subtraction
property of equality to write 6 2 = 3 2, or 4= 1, we have shown
that the statement "If 2=1, then 2 2 = I 2 " is true.
Of course, a general statement in the form of a conditional is of no
value in a specific situation if the hypothesis of the conditional is false
5.4 Conditional Statements in Proofs 201
in that situation. The truth of "If p, then (f does not by itself guarantee
tlie truth of either p or q. But the truth of the conditional and of the
hypothesis is a different story, as we shaU see.
Let us go back, then, to the three statements (A), (B), and (C) in
volving x and y. Suppose that (A) and (B) are both true. ITiat is, suppose
that the conditional and the hypothesis are both true. Then it follows
logically that the conclusion (C) is true. Check this with our example.
This is a most important concept in mathematical proofs. It means that
we can assert (C) after we have proved or know that both (A) and (B)
have been established On the other hand, it does not mean that (B)
follows from (A) and (C). (In our example, x = y does not follow from
x 2 = ry 2 , since we could also have x = — y<) In general,,
if a conditional and its hypothesis are both known
to be true, then the conclusion of the conditional
is also true.
More concisely, if we know that the following two statements have
been established;
L if p. then q
then we may conclude that q has been established. This means, in our
example, that if we know the following two statements are both true:
(A) if as = y, then x 2 = y 2
(B) x=y
then we can conclude x 2 = y 2 .
How are you to know when a conditional is true? In our example,
the conditional
if x = y, then x 2 = y 2
is a theorem that can be proved using the properties of numbers. In our
geometry, a conditional is accepted as being true if it is a postulate, a
previously proved theorem, or part of a definition.
Let us look at two more examples.
Example I We know that the conditional
if JB m CD, then AB = CD
is true. Why? Suppose that we also know or have been able to establish
that AB ^ CD. What can we conclude?
202 Congruence of Triangles Chapter 5
Fjxamph 2 We know that the conditional
if two angles are right angles, then they are congruent
is true. Why? Suppose that you are able to establish that Z A and LB
are right angles. What can you conclude? If, in connection with the
same conditional, you are able to establish that LA^ LB, can you
then conclude that £ A and Z B are right angles? Why?
EXERCISES 5.4
Write the theorems in Exercises 16 in "ifthen" form. State the hypothesis
and the conclusion of each.
1, Supplements of congruent angles are congruent.
2, Right angles are congruent
H. Vertical angles are congruent.
4. Two angles of a linear pair are supplementary.
5. The intersection of two convex sets of points is a convex set.
6. The interior of a triangle is a convex set.
In Exercises 710, a statement p and a statement q are given. In each exer
cise, write the truth value {'that is, true or false) (a) for the statement "If p,
then q" and (b) for the statement "If q t then p" The answers to Exercise 7
have been given as a sample,
% p: ml A = 90 and mLB = 90
q; LA^LB
(a) T (b) F
8. p
9. p
9
10. p
VB is the midray of Z AVC.
ZAVBsb LBVC
AMB and AM a MB^
M is the midpoint of 3B.
Z A and Z B arc supplementary angles.
(f LA and Z B are a linear pair of angles.
In Exercises 112G, certain given statements are to be accepted as true.
Then a conclusion is stated. In each exercise, state whether the conclusion
is true, false, or inconclusive (that is, not enough information is given to
decide whether the conclusion is true or false).
11, Gkmi: x + y = 16, x  y = 12.
Concision: x = 15,
12, Given: If there is not u cloud in the sky, then it is not raining.
There is not u cloud in the sky.
Conclusion: It is not raining.
5.5 Proof* in TwoColumn Form 203
13. Given: If iherc is not a cloud in the sky, then it is not raining.
It is not raining.
Conclusion: There is not a cloud in the sky.
14. Given; Yon urc a member of the team only if you obey the training
rules
Ken is a member of the team.
Conclusion: Ken obeys the training rules.
1 5. Given; You are a member of the teai 1 1 only if you obey the training
rules.
Bill obeys the training nilei.
Conclusicm: Bill is a member of the team.
16. Given: 2% + 3y = 12, x  y = 4
Conclusion: 3x + 2u = 16
17. Given: The intersection of two convex sets of points is a convex
set. S and T are convex sets of points. S f\ T = H.
Conclusion; R is a convex set.
18. Given: If a and b are numbers and if ab = 0, then a = or b = 0.
a and b arc numbers, ab = 0, and a =£ 0.
Conclusion: h =
19. (.'teen: If mLA = 30 and mZB s 00, then LA and Z B are
complementary angles.
ZA and LB are complementary angles.
Conclusion: mL A = 30 and fnZB= 60
20. Given: Linda will marry Joe only if he will buy her a new house.
Joe will buy Linda a new house.
Conclusion; Linda will marry Joe.
5.5 PROOFS IN TWOCOLUMN FORM
Students often ask, "What is a correct proof?" Unfortunately, there
is no simple answer to the question. Making correct proofs is something
that each of us learns by experience. A proof that may seem convincing
to you may not be at all convincing to another person with much less
experience in geometry than you. A deductive proof of a theorem is a
set of statements, one or several or many, that shows how the conclu
sion follows logically from the hypothesis. To make a good proof it is
important to think clearly about what is given and what is to be proved,
and to consider various possibilities of statements which will lead from
what is given to what is to be proved. It will help you considerably if
you have a firm understanding of the postulates, definitions, and
theorems already stated or proved.
The proof of a theorem is often given in paragraph form. We have
used this form of proof for most of the theorems proved thus far. For
illustrative purposes look at the proof for Theorem 4.13 in Chapter 4.
204 Congruence of Triangle* Chapter 5
THEOREM 4.13 Any two right angles are congruent
Proof: Every right angle has a measure of 90, Hence all right angles
have the same measure, and hence they are congruent to each other.
This proof consists of two sentences. Since the second sentence
consists of two parts, there arc really three steps in the proof. These
three steps, or links, form a chain of reasoning that shows how the
conclusion,
they are congruent,
follows logically from the hypothesis,
two angles are right angles.
In writing this proof we did not provide reasons to support these three
steps because we felt that the proof as given could be understood by
someone with your background in geometry. If we were trying to con
vince someone with less practice, it would be necessary to give addi
tional statements for justifying each of the statements in the proof.
One of the advantages of the paragraph type of proof is that it is
not always necessary to give the reasons for all the statements when
those reasons are obvious. However, when you give a proof in this
form, you should be prepared to fill in the reasons.
A twocolumn proof of a theorem consists of a chain of statements
written in one column with a supporting reason for each statement
written in a second column. When a statement in this chain is estab
lished because it is part of the hypothesis of the theorem, we simply
write "hypothesis" or * 'given" as the reason. Otherwise, a statement
may have as its supporting reason a combination of a conditional and
its hypothesis. As stated before, a conditional is acceptable if it fa a
postulate, a part of a definition, or a previously proved theorem. The
hypothesis of this conditional should have appeared as an earlier state
ment in tile proof. The conclusion of the conditional should apply to
the statement that is being supported
A twocolumn proof of Theorem 4.13 follows. Note that in reasons
which are conditionals, we write the numbers of the statements in
which we have established the hypothesis of the conditional.
THEOREM 4.13 Any two right angles are congruent.
RESTATEMENT; If Z A and L B are any two right angles, then
Proof:
Hypothesis; £ A and L B arc right angles.
Conclusion: £A ~ LB
5.5 Proofs in TwoColumn Form
205
1 . L A and Z B are right angles.
2 mLA = 90, mlB a 90
3. mZA = mZ£
4. IA^ LB
1. Hypothesis
2. If an angle is a right angle (1),
then its measure is 90,
3. Substitution properly of
equality (2)
4. If two angles have equal meas
ures (3), then they are con
gruent
We note several important points about this proof.
1. The proof is not complete until the last statement in the left
hand column is the same as the conclusion.
2. When a statement Is part of the hypothesis, we write "hypoth
esis" or "given" as its reason,
3. When a reason is in the "ifthen" form, its hypothesis refers to
an earlier statement or statements for support. For example, the
ifclause of reason 2 refers to statement 1, However, the then
clause of reason 2 refers to statement 2.
4. When a reason is not in die "ifthen" form, and it can he written
in that form, then it must satisfy the requirements stated in (3).
For example, reason 3 simply states: "Substitution property of
equality (2)," We could also have stated reason 3 as follows: "If
a, b t and c are numbers and if a = c and b = c (2), then a = h."
5. in proving the theorem we have not proved statement 4 con
sidered by itself as an isolated statement Rather we have proved
the following conditional: If statement 2, then statement 4.
Your teacher may permit you to list your reasons simply by identi
fying the postulate, definition, theorem, or property of equality which
supports each statement. If this is the case, the proof of Theorem 4.13
might read as follows:
Statement
1. L A and L B are right angles.
2. mlA = 90, m£B 90
3. mLA = mlB
4. £A==* IB
1. Given
2. Statement (1) and the defini
tion of right angle
3. Statement (2) and the substi
tution property of equality
4. Statement (3) and the defini
tion of congruent angles
206 Congruence of Triangles
Chapter 5
We have shown three examples of proofs of the same theorem.
Which is the best proof ? The answer is that they are all good. The two
column proof reminds us that we audi be able to give a reason for ever)'
statement we make, and it also makes it easier to see which hypothesis
we accept to begin our proof. Which proof you choose may depend on
whom you are trying to convince. Usually in writing a proof your ob
jective will be to convince your teacher that you understand the proof.
Your teacher may want you to have experience in writing both the par
agraph type of proof and the twocolumn type.
EXERCISES 5.5
1. In the following twocolumn proof several reasons are given in the
"ifthen" form. For each such reason indicate to which statement the
ifclause refers and to which Statement the thenclause refers.
THEOREM Supplements of congruent angles are congruent.
restatement; If /. < r i s a supplement of Z a and if L. d is a sup
plement of Lb and if La == Lb, then Lc m L<L
4
Proof;
Hypothesis; L c is a supplement of L a,
L d is a supplement of Lb.
Lam Lh
Conclusion: Lc ^ Ld
dXb
1. Z c is a supplement of L (t,
Ld is a supplement of Z b,
2. mLc+ mLa = ISO
mLd+mLb = 180
3. La== Lb
4. mLa = mLb
5. mLc + mLa = m/.d +
fflZ b
6. mLc = mLd
7. Lczz Ld
1. Hypothesis
2. If two angles are supplementary,
then the sum of their measures
is 180.
3. Hypothesis
4. Tf two angles are congruent, then
they have the same measure.
5. If x, y, and z are numbers and if
x = y and z = y, then x = z.
6. If a, b, x, y are numbers and if
a — b and x = «/, then x — a =
yb.
7. If two angles have the same meas
ure, then they are congruent.
5,5 Proofs in TwoColumn Form 207
2. Identity each of reasons 2, 4, 5 r 6, and 7 in Exercise 1 as a postulate,
definition, theorem, or property of equality.
3. Write a twocolumn proof for these theorems from Chapter 4,
;a) Complements of congruent angles are congruent.
In Vertical angles are congruent.
4* Write a proof for the following theorem in (a) paragraph form and
ib) twocolumn form.
THEOREM If two angles are both congruent and supplemen
tary, then each is a right angle.
5, Write a twocolumn proof of the theorem that vertical angles arc con
gruent using (a) the definition of vertical angles, (b) the definition of
a linear pair, (c) the theorem that if two angles form a linear pair, then
they are supplementary, and (d) the theorem that supplements of con
gruent angles are congruent (Hint: In the figure, let La and Z c he a
pair of vertical angles. Prove Z a s Zc>)
In Exercises 611, you are to perform some experiments with physical tri
angles. You will need a ruler, a protractor, and a compass. You are tv use
these experiments as a basis for formulating the next three postulates in
Section 5.6. Do the required constructions and measurements carefully and
answer all the questions before proceeding to the next section.
6. Construct a triangle, AAJ3C, in which AB = 2 in., AC = 1^ in,, and
m/.A = 50, Measure the remaining three parts (BC, /_!&,/.€) of your
constructed triangle and compare your measurements with those of
two or three of your classmates. Are they nearly the same?
7. Construct a triangle, ARST, in which HS = 2 in., m Z R  40, and
mZ S = (it). Measure the remaining three parts { Z T, RT t ST) of your
constructed triangle and compare your measurements with those of
some of your classmates. Are they nearly the same?
8. Construct a triangle, APQR, in which EQ = 5 cm., FR = 4.5 cm.,
and RQ = 6 cm. (You may need to use a compass for this construc
tion.) Measure the three angles of your constructed triangle and
compare your measurements with those of some of your classmates.
Are they nearly the same?
J). Construct a triangle, &DEF, in which mLD = 50, m£B = 60, and
mZF = 70. Measure the three sides of your constructed triangle in
centimeters. Compare your measurements with those of some of your
classmates. Are they nearly the same?
208 Congruence of Triangles
Chapter 5
10. Construct a triangle, &LMN, iii which m£L = 40, LM — 5 cm,, and
M\ ! = 3.5 cm. (You may need to use your compass again for tills con
struction,) Measure Z M, L N, and LN of your constructed triangle and
compare your measurements with those of some of your classmates. Are
they nearly the same?
11. Using only your ruler, construct a triangle, AADE, which has no two
of its sides congruent. Now construct AA'DE* (distinct from AADE)
such that AADE a A'D'F!. (You need not restrict yourself to using
only your ruler for this construction.) How many of the six parts of
AADE did you use in obtaining AA'D'E'? Could you have obtained
AA'D'E by using a set of parts different from those that you actually
used? What is the least number of congruent parts of AADE needed
to be sure that AA'D'E 1 is congruent to AADE?
5.6 THE CONGRUENCE POSTULATES FOR TRIANGLES
We asked you to perform certain constructions in Exercises 5.5.
We now wish to examine these con
structions in more detail, but first
we need some definitions. In A ABC
(Figure 54) we say that Z A is in
cluded by sides AC and "KB. Simi
larly, we say that side B€ is included
by angles Z B and Z C
Definition 5.2 An angle of a triangle is said to be included
by two sides of that triangle if the angle contains those sides,
A side of a triangle is said to be included by two angles of
that triangle if the endpoints of the side are the vertices of
those angles.
_ln A ABC, shown in Figure 54, which angle is included by sides
BC and BA? Which side is included by Z A and Z C? Were you able
to answer these last two questions without looking at the picture of
the triangle? Without looking at a pienre of ARST, state which angle
is included by sides ST and RT, Which side is included by Z R and Z S?
In Exercise 6 of Exercises 5.5, you should have concluded that all
triangles having the given parts are congruent; similarly for Exercises
7 and 8. When tins is true, we say that the three given parts determine
a triangle. In Exercise 6, the three given parts of the triangle were two
sides and the included angle. In Exercise 7, the three given parts were
two angles and the included side, and in Exercise 8 the three given
parts were the three sides. In Exercises 9 and 10 ? however, you should
have found that not all triangles having the three given parts are con
56 Congruence Postulates for Triangles 209
gruent. How many triangles of different size can be constructed if only
the measures of the three angles are given (assuming there is at least
one triangle with angles having these measures)? How many triangles
of different sizes can be constructed using the data of Exercise JO?
How did you construct AA'HE' in Exercise 1 1? Make a list of the
steps you used. Perhaps you have one of the following lists.
list I
1. Draw AU^
2. Draw LA' =
LA,
3. DrawA'Fs
4. Complete the
construction by
connecting R' and
j r 7withD ? F,
List2
1. Draw ZA's
LA
2, Draw A' if ss
AD.
3, Draw LU zs,
LD.
4. Complete the
constmction by
drawing the sides
ot LA' and LD 1
long enough.
List3
1. DrawATFss
AD.
2. Draw an arc with
A' as center and
A'Ef as radius.
3. Draw an arc with
17 as center and
D'E' as radius.
4. Complete the
construction by
connecting the
intersection of the
arcs to A* and U.
A' D' a* n A' d
For each list, the figure at the bottom shows what the construction
looks like just before it is completed.
Tjook at the first list. How many side measures are used? How
many angle measures? Is the angle between the two sides? This com
bination oitwo sides and the included angle is abbreviated by the sym
bol S.A.S.; the correspondence ADE + — ► A'UE' is referred to as an
S.A.S, congruence because we feel that this much information about
congruent pairs is enough to guarantee that all matched pairs of parts
are congruent. We make this conclusion formal in Postulate 23,
Look at the second list. What combination of measures is used?
This combination of two angles and the Included side is abbreviated
A.S.A.; the correspondence ADE* — >A'D'E' is referred to as an
A.S.A. congruence. We make this conclusion formal in Postulate 24,
Look at the third list. This combination of three sides is abbreviated
S.S.S., the correspondence ADE< — *A'D'E' is referred to as an
S.S.S. congruence. We make this conclusion formal in Postulate 25.
210 Congruence of Triangtes
Chapter 5
POSTULATE 23 (The S.A.S. Postulate) Let a onetoone cor
respondence between the vertices of two triangles (not necessarily dis
tinct) be given. If two sides and the included angle of the first triangle
are congruent, respectively, to the corresponding parts of the second
triangle, then the correspondence is a congruence. (Sec Figure 55.)
A ABC
fl D
ADEF hv the S.A.S. Postulate
Figure 55
POSTULATE 24 (Tkts A.S.A. Postulate) Let a onetoone cor
respondence between the vertices of two triangles (not necessarily dis
tinct) be given. If two angles and the included side of the first triangle
are congruent, respectively, to the corresponding parts of the second
triangle, then the correspondence is a congruence. (Sec Figure 56.)
A ABC = ADEF by the A.S.A, Postulate
Figure 56
POSTULATE 25 (The EMS Postulate) Let a onetoone cor
respondence between the vertices of two triangles (not necessarily dis
tinct) be given. If the three sides of the first triangle are congruent,
respectively, to the corresponding sides of the second triangle, then
the correspondence is a congruence. (See Figure 57.)
B D
A ABC * ADEF by the &&& Populate
Figure 57
5,6 Congruence Postulate* for Triingles 211
Note that there is no A.AA. Postulate (Figure 58) and no S.S.A.
Postulate (Figure 59). Also note how the results of Exercises 9 and 10
of Exercises 5.5 arc related to this statement.
B D
AABCgt ADEF
AAHC £ ADEF
Figure 59
Our experience with physical triangles suggests that it would be
proper to include an S. A. A. Postulate (Figure 510). Actually, we do not
need such a postulate. The statement that you might expect as a postu
late is, in fact., given as a theorem in a later chapter. We defer the proof
because we need not only the congruence postulates but also a postu
late about parallel lines, which appears later, before we can prove the
S.A.A. statement. Since it is easy to prove later, we have decided not
to adopt it formally as a postulate.
AAiJC := AJ>EF b> S.A.A.
Figure 510
The A.S.A. and S.S.S, Postulates can be proved as theorems once
the S.A.S. Postulate is assumed. The proofs are difficult, however, and
so we have adopted these statements as postulates in order to make
simpler the development of our geometry.
212 Congruence of Triangles
Chapter 5
EXERCISES 5.6
In Exercises 116, like markings on the triangles indicate congruent parts.
In each exercise, determine if a pair of triangles can be proved congruent.
If a congruence can be proved, write a triangle congruence (in the form
ABAC ss &FDE) and the postulate (S.A.S., A.S.A., S.S.S.) you would use
to prove it. Exercise 1 has been worked as a sample.
J.
ABAC at A FDE
by&A£.
h D
3.
i *i»
6. R
7.
5.6 Congruence Postulate! for Triangles 213
13, V
214 Congruence of Triangles
17. For A ABC and ADEF,
Chapter 5
■\B  EF,
I As: IE,
AC^ED.
Write a congruence between the two triangles. What postulate are you
using?
IS. For ARSUmd &GKL,
Which two sides must be proved congruent if
AJRl/5 == &LKG
by the A.SA, Postulate?
19. For AFQR aud AQSR,
FRsiSn and PQ^SQ,
Arc the triangles necessarily congruent? Why? Draw a figure,
20. For A ABC and A BCD,
IABC=* IDBC and LACB== LDCB.
Are the triangles necessarily congruent? Why? Draw a figure.
In Exercises 2130, two triangles appear to be congruent in the given figure.
Iu cadi exercise, certain information is given about the figure. Assume that
all points are coplanar and have the relative positions shown.
(a) Copy and mark each figure, as was done in Exercises 1  1 (i, to show
the given information.
(b) If the given information is sufficient to prove the triangles con
gruent, state a congruence between the triangles and the postulate
you would use (S. A.S., A.S.A., S.S.S.) to prove it. If the given infor
mation is insufficient, write "Insufficient."
£1. M is thcmidpoinl of FQ. 22. A~D  EC at D.
PR==QR D is the midpoint of EC.
5.6 Congruence Postulate? for Triangles 215
23. VA is the midray of IBVC.
m£VAC=mlVAB
27. LFW NEM
D is the midpoint of LM,
LF = ME
DF = DE
N
24. Consider onlv APQR and ASRQ.
P0 _L UK, SR 1 OT, and
25. Consider onlv ARTS and AR VS.
KT^RV
ST=sW
m, STV = ml$VT
mlRTV = mARVT
L D M
In Exercise 27, if, in addition
to the information given for
the figure, we also haveLA T =
MN, is AFNDatAEXD?
Why? (Hint: Use ND in your
proof. This segment exists
even though it is not shown in
the figure.)
NHM
U is the midpoint of GK.
IsG LGK
MR 1 GK
N
26. AED, DFC
B is the midpoint of AC.
BE 1 AD
RF 1 £ft
AE = CF
30, TR ±
VR ^ SR
ARB
216 Congruence of Triangles Chapter 5
5 J USING THE S,A.S. r A.S.A., AND S.S.S. POSTULATES
IN WRITING PROOFS
In this section and lire following one, you will be asked to write
your own proofs. In writing these proofs you will usually need to prove
one or more pairs of triangles congruent by using the S.A.S., A,S,A.„
and S.S.S. Postulates. Therefore, in planning your proof, you should
look for the opportunity to apply one of these postulates to some pair
of triangles. We illustrate with some examples.
Example 1 If M is the midpoint of AB and CD, then AC ^ ED.
In starling to construct a proof for this statement, we first draw a
figure which seems to fit the hypothesis.
c
Figure Ml q
Figure 511 shows M to he the midpoint of segments A~R and CD,
(Can you draw a different figure which shows the same information?)
We have marked AM and JIB with the same number of marks since, by
the definition of a midpoint, we know thai A^f s MB. CM and MD
have been marked alike for the same reason.
Before attempting to construct a proof of a theorem, it is helpful to
have a definite plan in mind. In this example, we want to prove that
two segments are congruent. We know that two segments are congru
ent if they are corresponding sides of congruent triangles. Our plan is
to prove A AMC = ABMD in Figure 511. For completeness the fig
ure is included as a part of the proof, as it should be.
Pmof:
Hypothesis: M is the midpoint of AB and CD.
Conclusion: AC = H75
(Plan; Prove AAMC £s &BMD.)
5.7 Using the S.A.S., A.S.A., and S.S.S. Postulates 217
Reason
1. M is the midpoint of AB and
CD.
2.ZMe*m. CM a W)
3. £AMC=z I BUD
4 AAMCj^ ABAID
5. AC ^ tf D
L Hypothesis
2. Definition of midpoint (1)
3. Vertical angles arc congruent
4. S.A.S. Postulate (2. 3)
5. If two triangles arc congruent
(4), then their corresponding
parts are congruent
Note that there are essentially the following five steps to writing a
geometric proof in twocolumn form.
L Draw a figure which seems to fit the hypothesis and, where pos
sible, mark on the figure the information given in the hypothesis.
2. State what is given (the hypothesis) expressed in terms of the
figure.
3* State what is to t>c proved (the conclusion) also expressed in
terms of the figure,
4, State a plan for the proof. The plan need not he expressed in
written form, but should be carefully thought through before at
tempting to write the proof.
5. Write out the proof in twocolumn form,
Now let us examine the proof for Example 1. You probably noticed
in the statement of Step 3 that we relied strongly on the figure for giv
ing us the information that L AMC and Z BMD are a pair of vertical
angles. If our figures are carefully drawn, we can rely on them to give
us correct information, However, we must always be prepared to he
able to justify any information that we take from a figure by postulates,
definitions, and theorems. In the figure for the example, we know that
points A, Af, B are eollincar and that points C, M, D are collmear
(Why?). We also know that when two lines intersect, vertical angles
are formed. Since the figure clearly shows all of this information, we
did not bother to establish in our proof that I. AMC and Z BMD arc
vertical angles.
Since this may be your first experience in writing geometric proofs
in twocolumn form, your teacher may want you to write a more com
plete proof than the one given. In other words, your teacher may not
want you to assume any information from a figure without establishing
this information in your proof. If this is the case, we give a second, more
complete proof of the theorem in Example 1 for your consideration.
2 IS Congruence of Triangles
Ifajjxithesis: M is the midpoint of AH and CD.
Conclusion: AC =k BD
Chapter5
1, M is the midpoint of A73 and
CD.
2, M is between A and B, and M
is between C and D.
3, Ml, M& and m£> m£> are two
pairs of opposite rays.
4, IAMC and ZIttfD are ver
tical angles.
5, LAMCm IBMD
e.A~M==MB,CM=*MD
7. AAMCss AJ3MD
8. A~€=±BD
1. Hypothesis
2. Definition of midpoint (1)
3. Definition of opposite rays (2)
'1 Definition of vertical angles
(3)
5. If two angles are vertical (4),
then they arc eongnient.
6. Definition of midpoint (1)
7. S.A.S. Postulate (5, 6)
8. If two triangles are congruent
(7), then their corresponding
parts arc congruent,
Note that some of the reasons are written in abbreviated form. For
example, reason 4 is ''Definition of vertical angles'* rather than the com
plete statement of this definition. Be quite certain that you know the
complete statement before using the abbreviated form in your proofs.
Example 2 % in quadrilateral A#CD, TLB m BC and AB a CD, then
la = ia
Hypothesis: AD a 1$C
31 as ED
Conclusion: LA^ LC
(Plan: We want to prove L A = L C by showing they are corre
l ■ nding angles of congruent triangles. But there are no triangles in our
figure. We therefore draw DB to show which triangles wc shall use/)
S.7 Using the S.A.S.. ASA, and S.S.S. Postulates 219
Note that the segment JDH in Figure 512 is dashed to distinguish
it from the parts of the figure given in the hypothesis.
Figure 51S
We call a segment such as DB an auxiliary segment. Thus an aux
iliary segment is a segment that is not a part of the figure given in the
hypothesis, but does exist by the definition of a segment and the Point
Line Postulate. Such segments should be drawn into your figure when
it is convenient to use them in the proof. We now continue with our
plan.
Since two sides of AABD are congruent to the corresponding sides
of ACDB, we would expect to prove these triangles congruent by ei
ther the S.A.S, Postulate or the S.S.S, Postulate. If we were to use
S.A.S., we would need ZA^ L C. But since this is what we are trying
to prove, we cannot use it as part of our proof That leaves the S.S.S.
Postulate. In AABD, the third side is BD and in A CDB t the third side
is DB. But BZ5 and DB are the same segment and are congruent by the
reflexive property of congruence. We can now write the proof.
1. 35 s CD
2. AD ^ CB
3. Segment BD exists.
4 ED~.T5B
5. AABD =* A CDS
a lAs IC
1. Hypothesis
2. Hypothesis
3. PointLine Postdate and the
definition of segment
4. The reflexive property of con
gruence for segments.
5. $.$£, Postulate (I. 2, 4)
6. If two triangles are congruent
(5), then their corresponding
parts are congruent.
Note that when we wrote AD ^CSin statement 2 and BD s DB
in statement 4, we were adhering to the correspondence ABD < — *
CDB for the two triangles. Of course, it would be correct to write
315 m EC In statement 2 and BD m BD in statement 4. But it helps
to keep tilings clear for both the writer of the proof and someone read
ing it if it is written the way we wrote it, keeping the order of the letters
consistent with their order in the congruence we wish to prove.
220 Congruence of Triangles
Chapter 5
Example 3 If points are as shown in Figure 513 and if AB & CB f
FE ?~ DF, and ~BE ^ ED. prove AF = FC.
How do we formulate a plan for
this proof? A plan for a proof can
often be formed by "working back
wards." That is, we begin with the
conclusion and Lry to work our way
back to the hypothesis, If these
steps can then be reversed, we have
a plan for our proof. This method
LonsisEs ui writing ^or cnizuung
information in two columns:
or; A c
Figure 313
I Can Prove
If 1 Can Prove
1. JTszFC
L (a) AE^CD?
(b) FEzzUF\/
2. JEseCO
2, AAB£ss ACBD?
a AABE=*ACBD
3, (a) AB s CB (S))/
(b) ZB^Z/?(A)v/
(c) £EsAD(S)V
read line 1 as follows: I can prove AF ^ FC if I can prove
(a) AE at UD and (b) FE ^ DF. (Note that if we know statements (a)
and iTj) of line 1 in the second column, then we ran deduce statement. I
in the first column from Corollary 3.4.3.) Because (b) is given in the
hypothesis, it is checked. Statement (a) is then to be considered. It is
brought down to line 2. We read line 2 as follows: "1 can prove AE =
<OD if 1 can prove A ABC ss ACBD" Since this triangle congruence
is not given, we bring it down to line 3. Line 3 is read as follows: l T can
prove A ABE ^ ACBD if I can prove (a) AB 3 CB, (b) IB == LB,
and (c) HK "^ ED." Since all three of these statements are given or can
easily be proved, they are checked. We can now write the proof by re
le order of the statements.
Proof:
Hypothesis: AH ^ CB
FE^UF
BE^ED
Conclusion: AF ss FC
Statement
1. BEsBD
2, ZBs IB
4. AABE = ACUD
5.AEaCD
6. FE a DF
7, AFaJU
5.7 UsingtKe S.A.S., ASA, and S.S.S. Postulate* 221
Reason
1. Hypothesis
2. Reflexive property of congru
ence for angles
3. Hypothesis
4. S.A.S. Postulate (1, 2, 3)
5. If two triangles arc congruent
(4), then their corresponding
parts are congruent,
6. Hypothesis
7. Corollaiy 3,43 (5, 6)
Example 4 In Figure 514, all points are coplanar.
Hypothesis:
C is in the interior of L ADE.
E is in the interior of L CDB.
W IDE
W is CD
Copy and complete the proof
that AE=zC!B.
{Plan: We plan to use ADE
/.ADEel LCDB by the Angle Measure Addition Postulate, and
DE 3: UB. We can then use SAS.)
Figure 514
* CDB. We can show ID == CD,
L KD _L DC, BD X DE 1. Hypothesis
2. ml ADC = 90, mlBDE = 90 2. [?] (1)
3. C is in the interior of Z ADE. 3. [?]
E is in the interior of L CDB.
4. ml ADE = 90 + mlCDE
mlCDB = 9() + mlCDE
5. mZAD£ = mZCDB
6. ZADEj^ZCDB
7. AD s= CD, DE s Dft
8. AAEEtt A CDB
9. 32 fi CB
4, Angle Measure Addition Pos
tulate (2, 3)
5, Substitution
equality (4)
& mo)
7*
8. {6, 7)
9. eke), m
property of
222 Congruence of Triangles
Chapter 5
EXERCISES 5.7
In Exercises 122, write twocolumn proofs. In each exercise, copy the fig
ure and mark on it the given congruences.
L In the figure, S is between P and Q, BS ^ PQ, and Zla L 2. Prove
that S is the midpoint of FQ.
Copy the following outline and supply the missing reasons, includ
ing the numbers of supporting statements.
Statement
Reason
1. ll^JJl
2, BS=RS
a RS ± PQ
4. jLHSPandARSQ are right
angles.
5. /.RSF = £RSQ
a APSil^ &QSB
7. PS == P
8. S is between F and Q.
9. S is the midpoint of P%).
2. In the figure, LARD and Z CBE
are vertic al an gles, £ is the mid
point of DE, and ID St LE.
Prove that B is the midpoint of
3. In the figure, point B is between
points A and C and point C is
between points B and D. Given
CJR ^ DS, Zls Z 2, and
AB m CD, copy and complete
the proof that Zr =* IS.
1. Hypothesis
3. S
4. If lines determined by two seg
ments arc perpendicular (pi),
then [7].
5. Any two right angles are [?]
7. If {[UK then [g.
am
9. HKSE)
5.7 Usng the S.A.S., A.S.A.. and S.S.S. Postulate* 223
1. B is between A and C, and
C is between B and D,
2. KB s CD
a EC*BC
7. AACrt a ABDS
1. Hypothesis
^ m
a ru
4. LengthAddition Theorem for
Segments (1, 2, 3)
''• 'ill
6. Hypothesis
8, OiiTCsponding parts of congru
ent triangles are congruent
(0D
4. In the figure, A is between C and D, B is be
tween C and K, K is (he midpoint of AB.
L\ = Z 2, and AC = BC. Copy and com
plete the proof that LACK =z ABCK,
L Zl a Z2
2. DAC and EBC
3. /, 1 and Z G&K form a lin
ear pair, and Z2 and
Z CBK form a linear pair.
4. Z CAIC is a supplement of
Z I, and Z CBK is a supple
ment of Z2.
5. ZCAKci ZCBK
0. K is the midpoint of XB.
7. [?]*»[?]
9. ttlsiBC
10. ACAK a AGBK
11. ZACK a ZfiCK
1.0
2, Hypothesis
3, Definition of linear pair (2)
4. If two angles form a linear pair
(3), then they are supplemen
tary.
5, The supplements of congruent
angles are \J]ih 4)«
6. rj]
7. Definition of midpoint (6)
8. Hypothesis
9. Definition of congruent seg
ments (8)
io, rr{[?],[?j,rT)
ILEKB)
Chapter 5
224 Congruence of Triangtes
5. In the figure, LR = RN and
LM = MW Prove that
mZL== m£N.
A. Hypothesis: For A ABC, 5? is the midray of L ABC intersecting AT:
in D. AC is the midray of I BAC intersecting EC in E. Points R, A, R r S
are collinear in tliat order. L HAD m ZSB&
Conclusion: A~E=*W)
(Flan: Show that IS ss IS, that ZABE== /BAD (supplements of
congruent angles); and that vilBAE = mlABD (halves of equals),
Then use the AAA. Postulate to prove AABE B A BAD.)
7. GStant DA = CB
CBXA^
To Prove: /L&m£C
8. Docs your proof for Exercise 7 de
pend on points A, B, C, and D being
coplanar?
ft Given: £lm Z2
Z3 = Z4
To Proce; PS = RQ and
mZS= mZQ p
10. c liven: D and K are between
A and B as shown hi
the figure,
zage« zjbcd
dc=?EC
lb Prone: AAC& =g A BCD
AEsSD
1 1. Use the same hypothesis and figure given in Exercise 10 and prove
(a) AACD m ABCE. (b) AS a TIE.
12, In the figure, E is in the interior of
Z BA C, ADE, Z EDO' cs Z EDB,
and BD s CU Prove that A~S is
Lhemidrayof ZRAC
5.7 Using tht S.A.S., A.S.A,, and S.S.S, Postulates
13. Given;
us
AB = CB
AD = CD
To Prove: l\= £2
[Hint: DrawBD.)
14. In the figure, R, S, 7', and N
are collinear in that order.
£BSV=± ZFJTvMSs W t
TTa VS.
From L?=LV
s
N
15. In the figure, AEC, DEB,
ZI» Z2, Z3ss Z4,and
Prove that £C and £15 bisect
each other at E.
16. In the figure, E is the midpoint
of J5F, H is in the interior of
Z DEC, G is in the interior of
Z FEW. Z DEG s Z FEW, and
TJEssGE. Prove that Z tf =* Z G
17* Given the situation of Exercise 16, prove in two different ways that
LDEIls* LFEG.
18. Given: Zl=* Z2
Af is the midpoint of RS
rt^st
Prove: Z3=*Z4
£26 Congruence of Triangles
Chapter 5
19. In thefigure^AD = BC, E is the midpoint of CD. and AE s BE, Prove
that AC m& BD.
A B
Copy and complete the statements in the fallowing plan.
I Can Prove If I Can Prove
1, AC s£ BD
2. A ADC's A BCD
1. AADCssS A BCD?
2. (a) [U[2]{S)v'
(b) ^=m{A)?
(c) DC=CD{S)>J
3. AEDAs A£CB?
4 (a) ££^[7]
( C ) [U[i](S)v
\ T ovv reverse the steps in the second column and complete the proof.
a LADC s /. BCD
4. A[?J & A[7]
20. In quadrilateral A BCD, M is the midpoint of AC and Z 1 Si Z2, Prove
that any segment which contains M
and has its endpoints in the inte
riors of CD and AB is bisected at M,
[Hint: Draw EF such that EMF,
AEB, and DFC. Now prove
that AEMAmt AFMC and hence
that EM = FA/ . i
21. CHAUJ^CE PROBLEM. 1 1 1 A ABC T AC SS ATI
Prove that the correspondence CAB *. — *
BACisa congruence using the S.A.S. Postulate.
Hence conclude that A CAB ^ ABAC Does
tills prove that IBsz Z C? Why? Does this
prove that if a triangle has two congruent
sides, then the angles opposite those sides are
congruent?
5.8 Isosceles Triangle* 227
22. CHALLEN'CF PROBLEM. Ill AH ST, ZSsZF.
Prove thul I he correspondence 5Tft « — * TSft
is a congruence using the A.S.A. Postulate.
Hence conclude _that ASTR m ATSfl, Does
this prove that R$ =s JIT? Why? Does ihis
prove that if a triangle has two congruent
angles, then the sides opposite those angles are
congruent?
5.8 ISOSCELES TRIANGLES
Each triangle shown in Figure 515 has at least two congruent sides.
Such triangles are called isosceles (from the Greek words iso and skelm
meaning "equivalent" and " legs," respectively'!.
Definition 5.3 An isosceles triangle is a triangle with (at
least) two congruent sides. If two sides are congruent, then
the remaining side is called the base. The angle opposite the
base is called the vertex angle. The two angles that are op
posite the congruent sides are called the base angles.
In A ABC of Figure 515, A B and ~KC are congruent sides. The base
is JJQ the vertex angle is Z A. and the base angles are Z B and Z C.
Name the base, vertex angle t and base angles of ADEF in Figure 515,
If a triangle has three congruent sides as does &PQR in Figure
515, then any side may be considered as a base of the t dangle. The
angle opposite a base is considered the vertex angle corresponding to
that base, and the angles that include a base are called the base angles
corresponding to that base.
228 Congruence of Triangles
Chapw 5
Definition 5.4 A triangle with three congruent sides is
called an equilateral triangle. A triangle with three congruent
angles is called an equiangular triangle.
If you worked Exercises 21 and 22 of Exercises 5.7, you already
proved the following two theorems.
THEOREM 5.1 (The Isosceles Triangle Theorem) The base an
gles of an isosceles triangle are congruent
restatement: In A ABC (Figure 516), AC Si A75.
Prove that ZDs LC.
Flgur* 516 U
(Plan: We will show that, for AABC, the correspondence
BAC * — ► CAB is a congruence using S.A.S., and hence LB = LC
by the definition of congruent triangles,)
Proof:
Statement
1. XCzzAB
2. ZAa LA
3. tt^XC
4. &BACm ACAB
5. LB^ LC
1. Hypothesis
2. Reflexive property of congru
ence for angles
3. Symmetric property of con
gruence for segments (1)
4. S.A,S. Postulate (1.2, 3)
5. Corresponding parts of con
gruent triangles (4) are con
gruent.
Theorem 5.1 implies the result which follows. Its proof is left as
an exercise.
5.8 Isoitttas Triangles 229
COROLLARY 5.1J If a triangle is equilateral, then it is
equiangular.
We know from Theorem 5.1 that if a triangle has a pair of congru
ent sides, then the angles opposite these sides are congruent. The eon
verse of Theorem 5.1 is also true, and we state it as our next theorem.
THEOREM 5.2 (Converse of the Isosceles Triangle Theorem)
If a triangle has two congruent angles, then the sides opposite these
angles are congruent and the triangle is isosceles.
bestatkmevt: In A DEF (Figure 517), LE^ LF, Prove that
DF = DE and hence that A DEF is isosceles,
F Figures 517
(Plan: We will show that, for A DEF, the correspondence
EFD < — ► FED is a congruence using A.S .A. and hence DF » DE
by the definition of congruent triangles,)
Fnwf:
1. ZEss IF
2. EF^TE
3. ZF» IE
4. AEFD BS AFED
5. DF=sDE
6. ADEF is isosceles,
1. Hypothesis
2. Reflexive property of congru
ence for segments
3. Symmetric property of con
gruence for angles (1)
4. A.S.A. Postulate {1, 2 f 3)
5* Corresponding parts of con
gruent triangles (4) are con
gruent.
6, Definition of isosceles triangle
(5)
The proof of the following corollary of Theorem 5.2 Is left as an
exercise.
COROLLARY 5.2.1 if a triangle is equiangular, then it is
eqi
2 30 Congruence of Triangles
Chapter 5
EXERCISES 5.H
h Is every equilateral triangle isosceles? Is every isosceles triangle
equilateral?
2. In the fipnjre. F is in th<' interior of
A ABC, AB =* AC and FB ss PC
Prove, without using congruent
triangles, that ZABPftZACP.
3. Write a twocolumn proof of Corollary 5.1,1.
4. Write a proof, in paragraph form, of Corollary' 5.2,1,
5. Prove Theorem 5.1 using the S.S.S, Postulate.
6. In the figure, A RST is an isosceles
triangle with vertex angle at H.
Give a proof different from that
given for Theorem 5.1 that base
angles of an isosceles triangle are
congruent (Hint: Let M be the
midpoint of ST and prove
7. Given: FQS, FRT, and
£IS3 Z2
l^ve: &QPR is isosceles,
8. Civen: ABD t ACE,
AH =s AT?, and
Pnwe: AflDCsACES
9, challenge problem. Did you use the Isosceles Triangle Theorem in
your proof of Exercise 8? If not, write a different proof, using the Isos
celes Triangle Theorem* If you did, write a different proof in which the
Isosceles Triangle Theorem is not used.
10. Given; A/)£F is isosceles with
vertex angle at D,
Prove: (ft) AGDH is isosceles,
(b) Z UGH =s I DUG
5.8 lio&celes Triangle* 231
D
K G
H F
1 1. If points A, B, C, A F, F have the
betweenness relations shown in
the figure, if A ABC is isosceles
with vertex angle at li, and if
A AFC £s isosceles with vertex an
gle at F* prove AD as EC (Phn:
Use the Isosceles Triangle The
orem to prove A A DC fe ACEA
by A.S.A.)
12. In convex quadrilateral ABCD,
JS^AT) and 55 a; W< (a)
Prove I ABC ^ I ADC widiout
proving any triangles congruent,
(bj Draw AC intersecting B~D at
E. Prove BE = Wand AC ± ED.
13w In the figure, Z 1 ^ Z 2, points B, I>, F„ C are collinear.
Prove that AABC is isosceles.
BD = EC
232 Congruence of Triangles
14. Given: ARST is equilateral.
Prove: ARST is equiangular. (Plan:
Show that the correspond
ence ilST « — » TRS is a
congruence by S.S.S, and
that IR=*IT&£S by
the definition of congruent
triangles.)
t5. Given: APQR is equilateral, with
A, R. and C the midpoints of
i 3 ^, ^H, and M, respec
tively.
Prove: AARC is equiangular.
Chapter 5
1 0, challkn Gg l'noiiLEM. We can prove the A . S. A. Post ill ate as a theorem
once we have assumed the S.A.S. Postulate. Complete the proof of the
following statement.
Given an A.S.A. correspondence
FAM < — » RSP
as indicated in the figures, if
ZFs Zfl,
IFAM== ZS,
prove that A FAX! ^ A RSP.
Proof: There is a point If on FM such that FM' ~~. HP. Why? (Note
that our figure shows ,\f and Sf to be different points. We will prove
that they are the same point.) Therefore AFAM* ss. A RSP by S.A.S
{Show this.) IFAM' S I RSP. Why? Therefore AA$' = A*f by the
Angle Construction Theorem, It follows that M f = M (Why?) and
AFAM m ARSF.
5,8 lsosc«le* Triangtes 233
17, ciiamEkce problem. We can prove the S.S.S. Postulate as a theorem
once we have assumed the 5.A.S. Postulate and proved the Isosceles
Triangle Theorem. Complete the proof of the following statement.
Given an SS.S. correspondence
ABC <h> PQH
as indicated in the figure, if
AC =r VH,
prove that A ABC a APQR.
1. AB = FQ t B~€~i §R r
Wmm
2. There is a point F on the
opposite side of An from C
such that ABAFr £P>
3. There is a po int D on AF
such thatAD^ PR.
t &ABD s\PQR
1. Given
2. Angle Construction Theorem
3. Segment Construction The
orem
4. 5.AS Postulate (1,2, 3)
Since C and D are on opposite sides of AB, CD intersects AB in a
point H. Our figure shows H to he between A and B. We could have,
however, A = il.avB = H, or A B U, or HAB, Complete the proof
for the case where AHB by using the Isosceles Triangle Theorem and
the S.A.S. Postulate to prove that A ABC st AABD, It will then follow
from step 4 and the transitive property of congruence for triangles thai
A ABC" = &.PQR* Draw a figure and complete the proof for the other
cases, that is, for A = 1L B = H, ABH, and IIAB,
234 Congruent* of Triangles
Chapter 5
5.9 MEDIANS AND PERPENDICULAR BISECTORS
In Figure 518, M is the midpoint
of side W of A ABC Segment AM is
called a median of A ABC. Since each
side of a triangle has exactly one mid
point, every triangle has exactly three
medians. Draw a triangle and its me
dians. What property do the medians
appear to have?
Figure .MS
Definition 5.5 A median of a triangle is a segment whose
endpoints arc a vertex of the triangle and the midpoint of the
side opposite that vertex.
THEOREM 5.3 The median to the base of
an isosceles triangle bisects the vertex angle
and is perpendicular to the base,
restatement: In A ABC (Figure 5l c r,
AB == AC and AM is the median to BC.
Prove: L AM is the midray of Z BAC.
2. AM 1 W.
Proof: Copy and complete the following proof.
1. 7&mM
2. ZBs= LC
3. AM is the median to EC
4* M is the midpoint of EC,
6. AABM^AACM
7. ZBAAf == /.CAM
8. Af is in the interior of Z BAC,
9. AA? is the midray of L BAC.
10. /.BMA= I CM A
11. IBMA and Z.CMA are a
linear pair.
12. Z BAM and Z CAM are right
angles.
13. AM _L BC
1. Hypothesis
2. \t}{\2})
3. Hypothesis
4.0(1
8. Theorem 4.11 (4)
9. Definition of midray (7, 8)
io. 03 (00)
11. Definition of linear pair (4)
12. Theorem 4.12(10, 11)
13. 0(0)
5,9 Medians and Perptndicular Bisector*
In Figure 520, A ABC is isosceles with vertex angle at A. Ray AG
is the midray of the vertex angle. It appears from the figure that AG
bisects BC and is perpendicular to EC at M. This brings us to our next
theorem.
235
Figure 520
THEOREM 5.4 The midray of the vertex angle of an isosceles
triangle bisects the base and is perpendicular to it.
Proof: In proving this theorem it is necessary to show that ray AG
intersects BS in a point M that is between B and C (as Figure 520 sug
gests), By the definition of midray, AG is between rays A/5 and AC. By
the definition of betweenness for rays, B and C are in opposite half
planes with edge AG. By the definition of opposite sides of a line, there
is a point Af of AG between B and C, Since M is between B and C, it
follows from Theorem 4.11 that M is in the interior of L BAC. Since
M is in the interior of L BAC and since it is a point of AG or of opp AG,
but not both, it follows from Theorem 4. 1 that M is a point of AG. The
rest of the proof of Theorem 5.4 is straightforward and is left as an
exercise.
In Theorem 5.4, it was required to prove that a certain ray was per
pendicular to a certain segment and that the ray bisected the segment
The line that contains the ray and that is in the same plane as the seg
ment is called the perpendicular bisector of the segment in that plane.
We state this formally in the following definition.
Definition 5.6 The perpendicular bisector of a segment in
a given plane is the line in that plane which is perpcndiculat
to the segment at its midpoint.
236 Congruence of Triangles
Chapter &
In Figure 521, line I is perpendicular to A73 at M t the midpoint of
A~B. Since I and AB determine exactly one plane (Why?), we say that /
is the perpendicular bisector of AB in that plane. We sometimes write
"I — bisA~B" for "/ is the perpendicular bisector of A~B."
M
Figure 521
Note that, in space, there is more than one line (How many?) that
is a perpendicular bisector of a given segment. However, in a given
plane the perpendicular bisector of a segment is unique, since a seg
ment has exactly one midpoint and, by Theorem 4.14, in a given plane
there is exacdy one line that is perpendicular to a given line at a given
point on the line. The next two theorems serve to characterize the set
of points in the perpendicular bisector of a segment,
THEOR&W 5.5 (The Perpendicular Bisector Theorem) If, in a
lhvcu plant* a. Pis a nnitil on the perpendlCMbl I fceGtOE of AHk tfagB
P is equidistant from the endpoints of AR
restatement: In Figure 522, P is a point on line / in a plane a.
I JL bisAB at M in a, l
Prove: PA  FB
Proof: If P = M, then PA = PB
by the definition of midpoint. If P
is any point in I different from M,
then &APM £= ABFM by S.A.S.
{show this); hence PA == ?B by the
definition of congruent triangles
and PA = PB by the definition of
congruent segments.
Figure 522
5.9 Medians and Perpendicular Bisectors 237
THEOREM 5,6 (Converse of the Perpendicular Bisector The
orem) If, in a gpven plane a , P is equidistant from the endpoints
of AB t then P lies on the perpendicular bisector of XB,
RESTATBMKNT! In plane a, PA s PB, M is the midpoint of AB.
Prove: P is on I, the perpendicular bisector of AB.
l
Proof; If P is on line AB, dien P = M tacause AB has only one mid
point, In this case, P is on line I by the definition of the perpendicular
bisector of a segmen t, I f P is not on line AB, then AAPM =s A BPM by
S.S.S. (show this). Therefore £AMP is a right angle (Why?) and PM
± bis AU. Since, in a given plane, a segment has only one perpendicu
lar bisector, P is on I,
EXERCISES 5.9
L Copy and complete the proof of Theorem 5.3.
2, Complete the proof of Theorem 5,4 by writing it in twocolumn form.
You may assume that it has been proved thai AG intersects J5C in a point
M which is between B and C, (See Figure 520.)
3. Copy A ABC below.
(a) Construct the median from A to ISC.
(b,i On the same figure construct
the midray of L BAC.
(c) Does the midray contain the
median?
{d) What must be true about
A ABC if the midray from A
is to coincide with the median
from A?
238 Congruence of Triangles
Chapter 5
4. Copy A DEF in the figure.
(a) Construct the perpendicular
bisector of EF in plane DEF',
(b) Does the perpendicular bi
sector of EF contain point D?
(c) What must be true about
ADEF if _the perpendicular
bisector of EF is to contain D?
5. From which theorem may we deduce that the vertex of the angle op
posite the base of an isosceles triangle lies on the perpendicular bisector
of the base?
0* In the figure. A, B, C, D are distinct coplanar points, KB St /03, and
(a) Is BD tile perpendicular bisector of AC ?
(b) Is AC the perpendicular bisecLor of BD?
(c) Which auxiliary segment is needed to
prove AABCm AADCf
(d) How do we know this segment exists?
(e) Why is IB^ID?
(f) If your answer to (a) is Yes, prove it is
correct with a paragraph style proof.
Do not use congruent triangles,
(g) If your answer to (b) is Yes, prove it is
correct with a paragraph style proof. Do
not use congruent triangles.
7. In the figure, / is the perpendicular bisector of FQ. If the lengths of .seg
ments are as marked, find a, h, c, and d.
12/ /
/ V
\J \
K 8
c \
rf%.
y^
5.9 Medians and Perpendicular Bisectors 239
I In the figure, A ABC is isosceles with AR = BC. If BD is a median,
prove that AABD s ACBD.
h In the figure, A PQR is isosceles with PQ b HI, The midrays of L QRP
and L PQR intersect at T. Prove thai P? 1 QR. Do not use congruent
triangles in your proof.
10. If t _L bis A~B at M, and if P is a point on / different from M, prove that
FKI is a median of A PAR.
11. If S is the midpoint of (JR andPS 1 pR, prove that APQR is isosceles.
(It. is not necessary to use congruent triangles in your proof,)
12. Given the figure as marked with F the midpoint of A~B, prove that
DFl IE
13. < itALLtvGt problem. If AB and ($R are coplanar and congruent seg
ments, n 1 btsA~Q, m 1. bis BR, and mC n = P, prove that AABP =*
AQRF.
240 Congruence of Triangles
Chapter 5
In Exercises 1422, you will need to prove certain triangle* congruent that
arc not necessarily coplanar. You should look at the figures in perspective
and be aware that angles or sides that are congruent may not look congruent
in the figures. In every case, carefully draw a figure on your own paper and
mark on it the congruent parts before writing a proof.
14. In the figure, points A, B, and C
are in plane a. Point P is not in
plane a. TB _ AB and FB 1 BC.
If L BAC = £BCA t prove A~P ss
UP,
15, In the figure, points D, E, and F
are in plane /?. Point R is not in
plane £ If RD jl DE f KD _ DF,
and DE^W, prove A REF is
isosceles and that LREF ^
£RFE.
16. In dihedral angle A RS H. AUis in plane a, BC is in plane /?, ZC ± iT5 r
HC i RS. D is in /IS. and \A('.B is isosceles with vertex stl C. Prove
that A BDA is isosceles.
5.9 Medians and Perpendicular Bisectors 241
17. Gives TR, W, and KT bisect each other at S. Prove AKDP =± AT/R
A
^— ^
18. In dihedral angle EABF, E is in plane ft, K is in plane /J, and AB con
tains C and D, If triangles DEC and DFX.' are isosceles with vertex an
gles at £ and F, respectively, prove that L EOF == Z ECF, (Plan: Prove
AEDF s AECF by S.S.S.) {See figure above at right)
19. In Exorcise 18, is DC l>etween rays DF and DE? Can we prove
Z.EDF& Z ECF by the Angle Measure Addition Postulate?
20. In the figure, PA = PB =
PC and lAPB=zlBPC=z
I A PC. Prove that AABCte
equilateral.
21. In Kxercise 20, if we accept the hypothesis (but disregard the figure}, is
it possible that P, A t B> C are coplanar? Try to draw a figure for the
"plane case,"
22. caiAjxENCE problem. In the fig]:. /' ;iru.l pare on opposite sides of
plane a which contains points D, E, and EUPQ ±DK at E,TQ ± FE
at E, and M == @E, prove &FDF ^ A QDF.
242 Congruence of Trfangles Chapter 5
CHAPTER SUMMARY
The Definitions
If t for A ABC and ADEF.
L A *. — » ZD, KB < — » DE,
IB < — * IE, W < — > EF,
I C * — ► IF, AC < — > JJF y
we indicate this correspondence by writing ABC <—— * DEF where the or
der in which the vertices arc named preserves the six correspondences. We
speak of the pairs In the six correspondences named as CORRESPONDING
PARTS of the two triangles. Two triangles (not necessarily distinct) arc
CONGRUENT if and only if there exists a onetoone correspondence be
tween their vertices in which the corresponding parts arc congruent Such
a onetoone correspondence between the vertices of two congruent trian
gles is called a CONGRUENCE.
An ANGLE of a triangle is said to be INCLUDED by two sides of that
triangle if the angle contains those sides. A SIDE of a triangle is said to be
I N( .1 .UDED by two angles of that triangle if the cudpoints of the side are
the vertices of those angles.
The combination TWO SIDES AND THE INCLUDED ANGLE is
abbreviated by the symbol S. A«S, If there exists a correspondence between
two triangles such that S.A.S. of one triangle are congruent to S.A.S, of the
second triangle, then we call this an S.A.S. CONGRUENCE,
The combination TWO ANGLES AND THE INCLUDED SIDE is
abbreviated by the symbol A.S.A. If there exists a correspondence between
two triangles such that ASA. of one triangle are congruent to ASA, of the
Second triangle, then we call this an ASA. CONGRUENCE.
The combination THREE SIDES is abbreviated by the symbol S.S.S.
If there exists a correspondence between two triangles such that S.S.S. of
one triangle are congruent to S.S.S. of the second triangle, then we call this
an S.S.S. CONGRUENCE.
If a triangle has (at least) two congruent sides, it is called an ISOS
CELES triangle. The VERTEX ANCLE of an isosceles triangle is the angle
included by the congruent sides. The BASE of an isosceles triangle is the
side that is opposite the vertex angle. The BASE ANGLES of an isosceles
triangle are the angles whose vertices are the endpolnts of the base. An
EQUILATERAL triangle is a triangle that has three congruent sides. An
EQUIANGULAR triangle is a triangle that has three congruent angles. A
MEDIAN of a triangle is a segment that has for its endpoints a vertex of the
triangle and the midpoint of the side opposite that vertex,
In a given plane, the PERPENDICULAR BISECTOR OF A SEG
MENT is the line that is perpendicular to the segment at its midpoint.
A statement of the form "If fh then </" is called a CONDITIONAL.
The statement p is called the HYPOTHESIS of the conditional and the
Chapter Summary 243
statement q is called the CONCLUSION. If a conditional and its hypothesis
are known to be true, then it follows logically that its conclusion is also true.
The statement **ff q, then p" is called the CONVERSE of the statement
"If p, then q" and the statement "If p t then q" is called the converse of the
statement "If q y then p" A definition in "ifthen" form is to be understood
as a conjunction of two statements "If p, then q" and "If q t then p" Some
times this is abbreviated to "p if and only iiq,"
The Postulates
There were three postulates in this chapter having to do with congru
ent triangles. We list them In their abbreviated form only. Be sure that you
know the complete statement of each postulate.
POSTULATE 23. The S,A<S. Postulate.
POSTULATE 24. The \SJi. Postulate,
POSTULATE 25. Tlie S.S.S. Postulate.
The Theorems
There were six tlieorems and two corollaries in this chapter. We list
them in the order in which they appeared. Re sure that you know what each
theorem says and that you understand its proof.
THEOREM 5 J (77ie hosceles Triangle Theorem) The base angles
of an isosceles triangle are congruent.
COR OLLAR Y 5.1.1 If a triangle is equilateral, then it is equiangular.
THEOREM 5.2 (Converse of the I*o$eek* Triangle Theorem). If a
triangle has two congruent angles, then the sides opposite these angles
arc congruent and the triangle is isosceles,
COROLLARY 5,2. 1 I! a Lrifuigle is equiangular, then it is equilateral
THEOREM 5.3 The median to the base of an isosceles triangle bi
sects the vertex angle and is perpendicular to the base*
THEOREM 5.4 The midray of the vertex angle of an isosceles tri
angle bisects the base and is perpendicular to it.
THEOREM 5.5 (The Perpendicular Bizector Theorem) If, in a
given plane a, P is a point on the perpendicular bisector of KB, then P
is equidistant from the endpofnts of KB.
THEOREM 5.6 (Converse of the Perpendicular Bisector Theorem)
If, in a given plane «. P is equidistant from the endpoints of KB, then
P H*s on die perpendicular bisector of KB.
244 Congruence of Triangles Chapter 5
REVIEW EXERCISES
1. Write the fallowing statement as a conjunction of two statements in the
"ifthen" form: A triangle is equilateral if and only if it is equiangular,
2. In your answer to Exercise 1, what is the second statement called with
respect to the first? Are both statements true?
3. Write the converse of the statement: If two triangles are congruent,
then their corresponding parts are congruent
4. In Exercise 3, are both the statement and its converse true?
5. Write the convene of the statement: Vertical angles are congruent.
6. In Exercise 5. is the given statement true? Is its converse true?
■ In Exercises 7 IS, decide if the sentence is true or false.
7. In the statement, "If p, then q" p is the conclusion and q is the
hypothesis.
8. In the statement, "p only if q," p is the conclusion and q is the
hypothesis,
9. In the statement, "p if tf," p is the conclusion and q is the hypothesis.
10. If we know the statement "If jt>, then q" is true and if we know that p
is true, then we can conclude that q is true.
11. TJie statement "If p, then qr" may be true even though the statements
p and q are both false,
12. If A is on the perpendicular bisector of CD, then AC = AD,
13. If, in a given plane, PE = PF, then P is on the perpendicular bisector of
EF.
14. In a given plane, more than one line can be drawn which is both per
pendicular to a segment and bisects the segment.
15. In a given plane, more than one line can be drawn which bisects a
segment.
16. If R, S t and T are not colLuiear and R is on the perpendicular bisector
of ST, then A RST is isosceles,
17. Congruence of triangles has the transitive property.
18. In our formal development of geometry, we postulated that the sides
Opposite a pair of congruent angles in a triangle are congruent.
19. Copy and complete: If, in a given plane, R is (?] from the endpoints of
ST. then R lies on the (7] [3 of ST.
20. Copy and complete: If JAB is a segment and if R and S are disti n ct points
such that R, S, A, H are coplanar, RA = RH, and M = SB, then jTj is
the 7]l] of A3
21. If, la A ABC, ED bi sects Z ABC and intersects AC at D, ~KB m EC,
prove that D in the midpoint of AC.
22, If, in the figure, A~B  BC, does
it follow that lBADs*£BCD
because base angles of an isosceles
triangle are congruent? Why?
23, In the figure for Exercise 22, if
■\r. w. I BAD a . BCD. and
D is in the interior of AABC t
write a plan for proving A ABD ^
ACfJD.
24, Given: AC^BC
lACD^ Z.BCE
ADEjyEB
Prove; CD rr CK
Revktw Exsrcises 245
B
25. Given:
I'rnt r;
ADC,
BEQ BPD
APE, 11? EM RE
LCDP& I CEP
AAFB is
isosceles.
26. Given: A3 = CD t AMQ
BXSD.aBdADssEC
Prove: (a) A ABD ss ACDB
(b) ^lsZ2
(c) A A CD ss A GAB
fd) Z3aZ4
(e) ACMD as AAJlfB
(f) MJs the midpoint of
BDandAT,
27. In the plane figure, \f is the
midpoint of segments A~C and
EP, BE sm Prove
(a) AAMFss&CME.
(b) AABC== ACDA.
28. Given: ADE, DEB
AD"? BE
CD s ce
Prove: A ABC is isosceles.
Prove, in order,
Zl = /2
Z3= i. J
AACDj^ ABCtf
AC m BC
A ABC is isosceles.
Chapter
Carl Stru wc/Monkmetjtfr
Inequalities
in Triangles
6.1 INTRODUCTION
In Chapters 3 T 4, and 5 we developed the concept of congruence
for segments, angles, and triangles. When we say that two segments
are congruent, we mean that two numbers associated with these seg
ments, their lengths, are the same. The concept of congruence for an
gles and for triangles is based also on the concept of equal measures,
and hence on the concept of equality for numbers.
In this chapter we are concerned with segments and angles, in par
ticular with segments that are not congruent to each other and with
angles that are not congruent to each other. In other words, wc are
concerned with segments of unequal measures and with angles of un
equal measures. In comparing two segments {or angles) that are not
congruent to each other, it is useful to know which one has the larger
measure. To express such a comparison it is convenient to use some fa
miliar words and symbols defined formally as follows.
Definition 6.1 XB > CD if and only if AB > CD< AB <
CD if arid only if Afl < CD.
Definition 6.2 I ABC > LDEFii and only if m I ABC >
mLDEF, IABC< IDEF if and onlv 'if mlABC<
mLDEE.
248 Inequalities in Triangles Chapter 6
When used to compare segments, the symbol "]> M may be read as
"greater than," or "longer than," or "larger than." Similarly, **<" may
he read as 'less than," or "shorter than," or "smaller than."
When used to compare angles, the symbol "]>" may be read as
"greater than" or "larger than." Similarly, "<*' may be read as "less
than" or "smaller than,"
Note that Definitions 6, 1 and 6.2 express a comparison of segments,
or of angles, in terms of an inequality involving numbers. It should be
clear that an inequality involving segments or angles is really a state
ment of "notcongruence" together with a statement of which seg
ment, or angle, has the greater, or lesser, measure. Because the proper
ties of inequalities for numbers are essential for the development of in
equalities for segments and angles, we review them in Section 6,2.
Tliis chapter includes several important theorems, some of which
involve comparisons of parts of one Lriangle. Others involve compari
sons of parts of one triangle with parts of another triangle. To save time
we shall, in some cases, give "abbreviated proofs" of these theorems,
that is, just the key steps in the proofs. You should be able to supply a
complete proof if asked to do so.
6.2 INEQUALITIES FOR NUMBERS
We begin by defining* formally, less than and greater than for
numbers.
Definition 6.3 If a and b are numbers, then a K h if and
only if there is a positive number p such that b — a + p.
Also a > b if and only if there is a positive number p such
that a = b + p.
Definition 6.4 If a and b are numbers, then a < b if and
only if a < b or a = b.
Example 1 The statement 4 < 5 is read "4 is less than or equal to 5"
and, by Definition 6.4, it means 4 < 5 or 4 = 5. We know that a dis
junction of two statements is true if either of the two statements is true.
Therefore 4 < 5 is a true statement because 4 < 5 is true. Similarly,
5 < 5 (which means 5<5or5 = 5)isa true statement because 5 = 5
is true.
Most of the numbers in our geometry are positive numbers and
represent lengths of segments, measures of angles, measures of plane
6.2 InequalJttot for Numbers 249
regions (areas), or measures of regions in space (volumes). From Defi
nition 6.3 we can conclude that if x, y, and * are positive numbers and
if x = if + z, then both tj and z are less than x. We get this by first con
sidering z as the positive numt>er p in the definition and then consider
ing y as the positive number p. This gives us Lhe two statements
x = y + p and x — z 4 p«
Thus, by Definition 6.3, y < x and z < x. That is, both numbers in a
sum of two positive numbers are less than the sum. For example,
15 = 9 + a
Hence 9 < 15 and 6 < 15. From the second part of Definition 6.3, we
can conclude that the sum of two positive numbers is greater than
either of them. For example,
32 = 21 J 11,
Heneo32>2J and 32 > 11.
The following theorem is easy to prove using Definition 6.3.
THEOREM 6.1 If x and y are numbers, then x < y if and only if
There are two parts to Theorem 6.1,
1 . If x and y are numbers, and if x < y, then y > x.
2. If x and y are numbers, and if y > x, then x < y.
Proof of part b
Statement Reason
1. x and y are numbers and 1. Hypothesis
x<y.
2. y = x + p, where p is a posi 2, Definition 6.3
tive number.
3. y > x 3. Statement 2 and Definition
6,3
JVoo/ of part 2: Assigned as an exercise.
We now state five properties of order (inequalities) that are helpful
in proving theorems about geometric inequalities. You may consider
these properties as postulates for the real number system, although we
could prove Properties 03, 04, and 05 by using Definition 6,3 and
Properties QJ and 02.
250 ln«qu*fties in Trianglss Chapter 6
01 The Positive Closure Property. If x and y are numbers, and
x > and y > S then x + t/ > and xy > 0.
02 Trichotomy Properly. If x and t/ are numbers, then exactly
one of the following is true: x < y> x = I/* x > j/,
03 Transitive Property. Tf a;, y, and z arc numliers, and if x <C J/
and ( < # then x < z. Also, if * > y and y > £, then % > %
04 Addition Property. If a, h, x, and y are numbers, and if
x <C ^ and a < fe, then i + fl < y t &• Also, if x ]> t/ and a > i>,
then x + a > y + fc.
05 Multiplication Property. If x, y, and z arc numbers and if
x < y and s > 0, then xz <C ys. Also, if x >■ y .and z > 0, then
We now prove several theorems which are useful in the sections
that follow.
THEOREM 6.2 AS > CD if and only if CD < "KB,
There are two parts to Theorem 6,2.
1. If AB > CD, then £75 < "KB.
2. If CD < AB r then IB > CD.
Proof of part 1:
Statement Reason
1. £B>CD L Hypothesis
2. AB > CD 2. Definition 6.1
a CD<AB 3. Theorem 6.1
4. CD < ZB 4. Definition 6. 1
Proof of part 2: Assigned as an exercise,
THEOREM 6.3 IABC> ADEF if and only If
IDEF<: I ABC.
Proof; Assigned as an exercise.
THEOREM 6.4 Let three distinct collinear points A, B, C be
given. Then ACB if and only if AB > AC and AB > BC
6.2 Inequalities for Kumbers 251
Tnere are two parts to Theorem 6.4. (Draw a figure showing the re
lationship between points A, B, and C.)
J, Given three distinct collinear points A, B ? C, if point C is be
tween points A and B, then AB > AC and AB > BC.
2. Given three distinct collinear points A, B, C, if AB > AC and
AB > BC^ then C is between A and B.
Proof of part ! . Since C is between A and B, AB = AC f BC, Why?
But AC and BC are positive, Hence by Definition 6.3 AB > AC and
AB > BC. %
Proof of part 2; Points A. B, C are coUinear (given), so exactly one of
them is between the other two. Why? That is, we must have exactly
one of the following three belweenncss relations: BAC t ABC, or
ACB. We shall show that the first two of these are impossible; hence
the only remaining possibility is ACB. Suppose that A is l>etween B
and C, Then, by the first part of Theorem 6,4, we must have BCy AB.
But this contradicts the hypothesis that AB > BC; hence A is not be
tween B and C. Now suppose B is between A and C. Again, by part 1
of Theorem 9.4, we must have AC > AB and this contradicts the hy
pothesis that AB > AC; hence B is not between A and C. The only
remaining possibility is that C is between A and B. This completes the
proof.
THEOREM 6.5 If point D is in the interior of I ABC, then
mlABC>mlABD and m£ABC> m£DBC.
Proof: Since D is in the interior of /.ABQ ray BD is between rays
BA andBC. By the Angle Measure Addition Postulate,
mLABC = mlABD + m£DBC>
Since mLABD and mLDBC are positive, it follows from Definition
6,3 that
m Z ABC > m Z ABO and m Z ABC > m Z DBC.
EXERCISES 6.2
In Exercises 110, identify the order property that is illustrated.
L If AB < 6. then AB # a
2. If a  fe < 15 and ft < 3, then a < 18,
3. If x < 7 and 7 < y, then x < y.
252 Inequalities in Triangles
Chapter 6
4. If a < 5, ihen 4o < 20.
5 If AB < RS and BC < ST. then AS + BC < HS + ST,
6. If Jm Z ABC > ^m Z RST, then m Z ABC> m £ R ST.
7. If Ali > CD and CD a EF, then AB > EF.
8. If x + 3 < 8, then a: < 5.
9. If x  y > 12 and y = 7, then x > 19.
10. If 3 > and 2 > 0, then 2 + 3 > 0.
11, in the figure, AD > BE and DC > EC.
Prove AC >' BC.
12. Given the figure for Exercise 11, if
m £ 1< m £2 and m Z3 < to Z4
prove that m Z ABC > m £ BAC.
13, Ii m/.A = 90 + k, where k > 0, and ZB is u supplement of ZA,
prove that Z 2? is an acute angle.
14. In the figure, £D =z £ DEC. 15. In the figure, CD JL bis AB and
Prove that m Z ABC > ffl Z D. CPB. Prove that AC > CE.
16. Prove part 2 of Theorem 6.1,
17. Prove part 2 of Theorem 6.2.
18. Prove Theorem 6.3.
19. Explain why Theorem 6.4 has the following consequence. If A, B, C
are three distinct collinear points, then C is between A and B if and only
if AC <A£ and BC< AB.
20. Explain why Theorem 6.5 has the following consequence. If D is a point
in the interior of /ABC, then
m I ABD < to £ ABC and m £DBC < m £ABC.
P
21. In the figure, M is the midpoint
of AT and BC. Prove tliul
m£BCD> ml B.
6.2 Inequality for Number* 253
22. For each figure use your protractor to measure Z BCD y LA, and L B,
Record your results in a table. How does m/ BCD compare with mLA
in every cu.se? m Z BCD with m L B in every ease?
m BCD
m A
mLB
254 Inequalities in Triangfes
Chapter 6
23. challenge problem. Gi vet i the coplanar potato A,B,C, D, J#, P such
that A, B, C are noncollincar, A~C~D, BMC, and AMP, prove that
P is in the interior of L BCD. (Hint: You must show that P is on the
tfside of S/5 and that P is on the Dside of BC.)
24, challenge pbohlem. Prove the Transitive Property of Order (Prop
erty 03). {Hint: You will nocd to use Definitions 8.3 and 6.4 and the
Addition Property of Kquality in your proof.)
6.3 THE EXTERIOR ANGLE THEOREM
In both figures of Figure 61, ZABC, ZBGA, and LCAB are
called interior angles of A ABC, We call /.BCD, which is adjacent
to LACB and forms a linear pair with it, an exterior angle of A ABC.
Both Z A and Z B are called nonadjacent interior angles of the exterior
angle Z BCD. The term "interior angle" is convenient when you want
to emphasize the distinction between an exterior angle of a triangle and
an angle of a triangle. \ T ote diat the adjectives "adjacent" and "non
adjaeerit" apply to an interior angle and descrrt>e its relation to a par
ticular exterior angle. We formalin these ideas in die following
definition.
Definition 6.5 Each angle of a triangle is called an interior
angle of the triangle. An angle which forms a linear pair with
an interior angle of a triangle is ealled an exterior angle of the
triangle. Each exterior angle is said to be adjacent to the in
terior angle with which it forms a linear pair and nonadjacent
to the other two interior angles of the triangle.
Every triangle has six exterior angles, two at each vertex, as shown
in Figure 62, The two exterior angles at each vertex are vertical angles
and hence are congruent.
6,3 The Exterior Angle Theorem 255
Figure fi2
Note that L DCE in Figure 62 Is not an exterior angle. Why? In
Figure 62 T L KAC and I HAB are the two exterior angles at vertex A
of &ABQ and LABC and LBCA are the two nonadjacent interior
angles of each of them. Name the two exterior angles at vertex B of
A ABC in Figure 62 and their nonadjacent interior angles.
If you worked Exercise 22 of Exercises 6,2, your results should sug
gest the following theorem.
THEOREM 6.6 {The Exterior Angle Tfteorem) Each exterior
angle of a triangle is greater than either of its nonadjacent interior
angles.
Proof: Let the vertices of the triangle be A . B, and C. T iet D he a point
on AC such that C is be n p
tween A and D. (See Figure /S. ""/
63.; We must prove that / X.
LBCO> LBAC
and that
LRCD> LB.
We first prove that j**
IBCDy LB. Figured
Let M be the midpoint of BC and let P be the point on AM such that
AMF and AM = MR Then AAMB == A PMC by S.A.S. (show this)
and mLBCP = mLB. Why? Since P is in the interior of /BCD,
m L BCD > m L BCP by Theorem 6.5. We have shown that
mLBCD > mLBCP
and that
mLBCP mLB.
It follows from the Transitive Property (03) that mL BCD > mLB,
and from Definition 6.2 we have LBCD > / B.
256 Inequalities In Triangles Chapters
To prove that Z BCD > Z BAG, we use the midpoint of AC and
show that the other exterior angle at C {LACE in Figure 64) is greater
than / BAG in the same way as in the part above. Since the two ex
terior angles at C are congruent (Why?), it follows that
IBCD> I BAG
Figure <M
The proof that we have given for one exterior angle of the triangle
can be easily modified to show that die theorem holds for any of the six
exterior angles of the triangle. However, it is not necessary to go to all
this trouble since our choice of the exterior angle at C was stricdy an
arbitrary choice. The fact that we were able to prove the theorem by
choosing arbitrarily any one of the six exterior angles of A A /?C insures
us that there is no need to prove the theorem for every exterior angle.
In the proof of Theorem 6,6\ we stated that point P (in Figure 6^3)
is in the interior of Z BCD. You are asked to prove this in the Exercises
at the end of this section.
COROLLARY 6.6, 1 If one of the angles of a triangle is a right
angle, then the other two angles of the triangle are acute angles.
Frvof: Assigned as an exercise.
COROLLARY 6.6,2 If one of the angles of a triangle is an obtuse
angle, then the other two angles of the triangle arc acute angles.
Proof: Assigned as an exercise.
It follows from the Exterior Angle Theorem (more directly from
its corollaries, Corollary 6.0.1 and Corollary 6.6.2) that no triangle has
more than one right angle or more than one obtuse angle. An important
kind of a triangle is one that has one right angle. There are special
names for triangles with a right angle and for triangles with an obtuse
angle.
6.3 The Exterior Angle Theorem 257
Definition 6.6 A right triangle is a triangle with one right
angle. The hypotenuse of a right triangle is the side opposite
the right angle. The other two sides of a right triangle are
called legs.
Definition 6.7 An obtuse triangle is a triangle with one ob
tuse angle.
Definition 6.8 An acute triangle is a triangle with three
acute angles.
Theorem 4.14 asserts that for each point on a line in a plane, there
is one and only one line which (1) lies in the given plane. (2) contains
the given point, and (3) is perpendicular to the given line. We can now
use the Exterior Angle Theorem to prove a companion theorem.
THEOREM 6, 7 Given a line and a point not on the line, there is
one and only one line which contains the given point and which is
perpendicular to the given line.
Proof: Let I be the given line and P the given point not on L In part 1
of the proof we show that there is a line containing P and perpendic
ular to L In part 2 we show there cannot be two such lines.
1. Existence. (See Figure 65,) Let A and B be any two points of
line I. Then PA is either perpendicular to / or not perpendicular to /.
If FA ± h then the existence part of our proof is complete. If PA is not
perpendicular to /, then there is a ray AC* with C on the opposite side
of / from P such that I PAB ==
ABAC. Why? There is a point D
on AC such that AD at Aft Why?
P and C are on opposite sides of /,
and C and D arc on the same side
of l\ hence P and D are on opposite
sides of /. Therefore PD intersects
/ at some point F. APAF^ &DAF
by SJLS. (shew tills} and so
L PEA as LDFA. Therefore, by
Theorem 4.12, LPFA is a right
angle and PF ± I
Figure &S
258 Inequalities in Triangles
Chapter 6
2. Uniqueness. (See Figure 6*6.) Let C be any point of / other than
F and let H be any point of / such that GPH. Then Z PCFis an angle
of APGF and is a nonadjacent *
interior angle of the exterior
angle Z PFIL Since m L PFH =
90, it follows from the Exterior
Angle Theorem that mlPGF
< 90 and therefore PG is not
perpendicular to L It follows
that PP is the only hue through
P and perpendicular to L
Figure fifi
EXERCISES 6.3
1. Refer to Figure 64 and prove that Z BCD > Z MCj thus completing
the proof of Theorem 6,6.
2. Prove Corollary 0.0. L
3. Prove Corollary 6.0.2.
4. Given A ABC with BCD and mLACD = 70, what must be true
about m LABC? About r« Z BAG? About mAACB?
5. In the figure, name the two
nonadjacent interior angles of
ZBAP "Which exterior angle
has A CAB and I ABC as its
nonadjacent interior angles?
Name all the exterior angles
shown in the figure and their
corresponding nonadjacent inte
rior angles.
Using the figure, copy Exercises 6 1 1 and replace the question marks by
<, = t or > to make a true statement, a, b t c t x t y._ z denote angle measures.
6.3 The Exterior Angle Theorem 259
6. If b = 40, then x [?] 40 and y [?] 140,
7. If c = 90, then z \t] 90, a Tj 90, b Q] 90, x (?)90, and y 7]90.
8. If a = 40 and b = 60, then 2 f7] GO.
9. If (/ = 140, then a [T] 140 and c [7 140.
10. If x = 130 t then h [?] 130.
1L If a = 55 and c = 80, then i/ [?] 80.
Refer to the figure for Exercises 1218. In each exercise, arrange the nuni
bers in order, starting with the smallest. In these exercises, <i, h t c> d, e,f t g
denote angles.
12. m La, m Z c
13. mLh t mLd
14. mle, mZc
15. mZb» BlZ/
16. m £c ml_a t mLe
17. mZc, m Z e, rn Z g
18. mZf ( mZa,»nZg 1 »nZ«
19. In the Bgure, prove that mlACB > mLB.
A
20. List the angles marked in the figure In Increasing order of size.
260 Inequalities in Triangles
Chapter6
21 ♦ In the proof of Theorem 6,B it was stated that the point P (Figure
was in the interior of L BCD, To prove this* wc must show that P is on
tile Dsidc of S3 and on the Bside of c3.
A C
Copy and supply the missing reasons iu the following proof.
Statement
ACD, AMP, BMC
M and F are on the same
side of S3.
M and B are on the same
side of 53.
P is on the Bside of CD.
A and D are on opposite
sides of BC.
A and P are on opposite
sides of BC,
P is on the Dsidt of BC,
P is in the interior of
/ BCD.
L Given
2, Theorem 2.2
3. m
4. Statements 2 and 3
5. \?}
7, Statements 5 and
8. EKHHIl)
22. Is it possible for a triangle to have two right angles? Justify your answer.
23. Suppose that I is a line in plane a and that F is a point not in a. Docs
Theorem 6.7 still apply? Draw a figure and explain your answer.
24. Given: AAMC f AFS> 25. Given: AADB.ACD,
FSC AD = "KB
Pmw,: Z1>Z2 Prove: LACW> LDBA
6.3 The Exterior Angle Theorem 261
h Cwtmt B is the midpoint of
333, 8 is the midpoint of FC,
Bcn
Prom: IDCE> LA
27. Gtem: ARTS, RWl\ SW
is the midray of L RST
Prove; Z3>Z1
28, Given: Quadrilateral A BCD,
BKC, 3D ss DE t L 1 ~
Z2
Prvce: Z3>Z1
29. Gkert: F is the midpoint of
AD and BE
Prove: Z2<Z1
30, Measure the sides in centimeters and the dingles in degrees of the scalene
triangle in the figure. Record the measurements to the nearest tenth of
a centimeter and the nearest degree in a table as shown. Yon will need
to refer to the results of this exercise in Section 6.4.
Angles
c = f?l
m/.A=\J\
ifiZB = f7
262 Inequalities m Triangles Chapter 6
31. Draw three scalene triangles of different sizes and shapes and label the
vertices and sides as in Exercise 30. Also make and record measure
ments for each triangle. You will need to refer to the results of this ex
ercise in Section 6,4.
32. a r allence problem , If Pis any point in the interior of A A 8C, prove
that m£APB> mlC.
33. challenge pkohlem. Prove that the sum of the measures of any two
angles of a triangle is less than ISO,
6.4 INEQUALITIES INVOLVING TRIANGLES
If all three sides of a triangle are congruent, then die three angles
of the triangle are congruent, and if two sides of a triangle arc congru
ent* then the angles opposite these sides are congruent. In this section
we investigate how the angles of a triangle are related to each other
when they are opposite Sides of unequal length.
Refer to the table you prepared in Exercise 30 of Exercises 6.3 and
answer the following questions.
Which side is the longest?
Which angle is the largest"
How are the longest side and the largest angle of
AABC situated with respect to each other?
Which side is the shortest?
Which angle is the smallest?
How are the shortest side and the smallest angle of
AABC situated with respect to each other?
Observe the order relation among the measures of the angles of
AABC and complete the following statement;
mL\2] > mZQ] > mZT].
Observe the order relation among the measures of the sides of
AABC and complete the following statement:
EJ>E>[&
How do the order relations among the measures of the angles of
the triangle compare with the order relations among the measures of
tile corresponding opposite sides of the triangle?
Refer to the tables you prepared for the three triangles in Exercise
31 of Exercises 6,3 and answer the same questions as the preceding
ones for each of these triangles. Make a general statement about the
relative position of Lhe longest side and the largest angle of a triangle;
the shortest side and the smallest angle.
6,4 Inequalities Involving Triangle* 263
The comparisons suggested by the preceding experiment are for
mulated in the following two theorems,
THEOREM 6*8 {AngleComparison Theorem) If two sides of u
triangle are not congruent, then the angles opposite them are not
congruent and the greater angle lies opposite the greater side.
Proof:
Given; AABC,
with AB > ~KC
To From: I ACS > I ABC (See Figure 67.)
Statement
1 „ AB > £C
2. AByAC
3. There is a point D on
such that KD zzlC.
4. AD = AC
5. AB>AD
6. D is between A and B.
7. D is in the interior of £.ACB.
8. mlACB>mAACD
9. lACDss I ADC
10. mlACD = mlADC
11. nZACB > mZ40C
12. ZADC> ZAJ3C
13. m£ADC>mLABC
14. mZACB>mZABC
15. lACBy IABC
Reason
1. Given
2. Definition 6. 1
3. Segment Construction The
orem
4. Definition of congruence for
segments (3)
5. Substitution Property of
Equality (2, 4)
6. Theorem 6,4
7. Theorem 4,11
8. Theorem 6.5
9. Isosceles Triangle Theorem
(3)
10. Definition of congruence for
angles (9)
11. Transitive Property (8, 10)
12. Exterior Angle Theorem
13. Definition 6.2 (12)
14. Transitive Property (U, 13)
15. Definition 6.2 (14)
264 Inequalities in Triages Chapter 6
We now state and prove the converse of Theorem 6.8.
THEOREM 6.9 (SideComparison TJieorem) If two angles of a
triangle are not congruent, then the sides opposite them are not
congruent and the greater side lies opposite the greater angle,
Proofs Let A ABC be any triangle with two angles that are not con
gruent Suppose it has been named so that these noncongruent angles
are / B and Z C and such that LC> L B, In terms of this situation
the hypothesis and the conclusion to be proved are as follows.
Hypothesis: LC> LB
Conclusion: AB > AC
(See Figure 6S.)
Since AB and AC are numbers, one of the following must hold:
(1) AB<AC
(2) AB = AC
(3) AB > AC
Which property of numbers are we using here?
The method of proof is to show that (1) and (2) are impossible, so
(3) must hold, thus proving the theorem.
(1) If AB < AC, then, by Theorem 6.8, LC < LB, This contra
dicts the hypothesis; hence AB < AC is impossible.
(2) If AB = AC then A ABC is isosceles and LC ss LB. Again
this contradicts the hypothesis; thus we see that AB = AC is
impossible.
It follows then that AB > AC so that AB > KC and the desired
conclusion has been proved.
The following two corollaries follow immediately from Theorem
6.9 and Corollary" 6.6.1.
COR OLLAMY 6.9. 1 The hypotenuse of a right triangle is the
longest side of the triangle.
Proof: Assigned as an exercise.
6,4 Inequalities Involving Triangles 265
COROLLARY 6.9,2 The shortest segment from a point to a line
not containing the point is the segment perpendicular to the line.
restatkmeoti Given a Hue I
and a point P that is not
on I, if PA J. I at A and
B is any other point of ?,
then PA < FB. (See Fig
ure 69.)
►J
Proof: Assigned as an exercise.
a B
Figure 69
When we speak of the distance between a point P and ft line l> we
naturally mean the shortest distance from J' to I. It follows from Corol
lary 6,9.2 that there is such a shortest distance, and so we make the
following definition.
Definition 6,9 The distance between a point and a line not
containing the point is the length of the perpendicular seg
ment joining the point to the line. The distance between a
line and a point on the line is defined to be zero.
It is customary to associate three "distances between ft point and a
line" with every triangle. With A ABC there is associated the distance
between A and EC, the distance between B and CA, and the distance
between C and AS. Any side (or its length) of a triangle may be thought
of as the base. Associated with each base is the segment (or its length)
joining the opposite vertex to a point of the line containing the base.
The following definition has two parts. In (I), we define base and
altitude, thought of as segments. In (2), we define base and altitude,
thought of as numbers (lengths of segments or distances).
Definition 6,10
L Any side of a triangle is a base of that triangle. Given a
base of a triangle, the segment joining the Opposite vertex to
a point of the line containing its base, and perpendicular to
the line containing the base, is the altitude corresponding to
tluit base.
%. The length of any side of a triangle is a base of that tri
angle. The distance between the opposite vertex and the line
containing that side is the corresponding altitude
266 Inequalities in Triangle*
Chapter 6
Figure 610 shows two triangles, A ABC and AA'B'C. In each tri
angle the segments from the vertices to the lines containing the oppo
site sides have been drawn. These segments are the altitudes of the
triangles. Thus XD is an altitude of A ABC. It is the altitude from A to
sideTJC. The point D is the foot of the altitude from A to EC. Note for
an acute triangle, such as AABC in the figure, that the foot of each
altitude is an interior point of a side. Note for an obtuse triangle, such
as A A'B'C in the figure, that the feet of two altitudes ar« • not points of
the triangle. Even though point E\ for example, is not a point of side
A'C. it is frequendy called the foot of the altitude from B' to side A^U.
Figure 640
Draw a right triangle, A ABC, with the right angle at C. Draw the
altitude from C to the hypotenuse. Is the foot of this altitude a point
of the triangle? Name the other two altitudes of A ABC. Are the feet
of these altitudes points of the triangle?
In Chapter 3 we postulated that if A, B, C are noncolliuear points,
then for distances in any system, AB \ BC > AC. The postulate was
called the Triangle Inequality Postulate. We now have the necessary
geometric properties to prove this statement as a theorem.
THEOREM 6.10 (Triangle Inequality Theorem) The sum of the
lengths of any two sides of a triangle is greater than the length of
the third side.
Proof: Given any triangle, it follows from the Trichotomy Property
for numbers that there is one side of the triangle which is at least as long
as each of the other two sides. Suppose, in A ABC, that BC j> AB
that BC > AC, (See Figure 611.)
Figure 611 B
6.4 Inequalities Involving Triangles 267
We must prove the following three statements:
(1) AB + BC>AC
(2) AC +BOAB
(3) AB + AC > BC
(1) By hypothesis, BC > AC. Since AB > T (that is, AB is a
positive number), we have
AB + BC>AC
by the Addition Property of Order (04).
(2) By hypothesis, BC > AB. Since AC>O t we have
AC + BC > AB. Why?
(3) On opp AC choose point D so thai AD = AB. (See Figure
612.)
C Figure G12
Since A is between C and IX A is in the interior of L DBC. Also
mLDBOmlABD by Theorem 65. But r»ZABD =
mLADB by the Isosceles Triangle Theorem, and so
mlDBOmlADB.
"Therefore, by the SideComparison Theorem (applied to
A DBC),
DC>BC
Since DC = DA + AC (Why?) and DA = AB, we have
DC = AB 4 AC
We have shown that DC > BC and that DC = AB + AC.
Therefore, by the Substitution Property,
AM + AC>BC
It follows from Tneorem 6.10 that, in informal language, the short
est distance between two points is the length of the segment joining
them. In formal geometry, of course, once a unit of distance is given,
there is only one distance between two points.
It follows from Theorems 6.8 and 6.9 that, in any one triangle, the
greater angle lies opposite the greater side and conversely, the greater
side lies opposite the greater angle. Let us now consider two compan
ion theorems involving two triangles.
268 Inequalities In Triangles Chapter 6
In Figure 613 t AB = A'W and EC = BC. What can you say
about these triangles if m LB=mL B'? How do AC and A 'C compare?
B a*
Fitfirt613 A C A'
In Figure 614, we again have AB = AB' and BC = B'C\ but this
time m L B > m L B\ Is the correspondence ABC < — *■ A'BC/ a con
gruence? I Tow do AC and A'O compare in this case?
B B'
A A'
Flip j re 614
In Figure 615, we have AB = A f B\
m£B<C.ni£W t Ts the correspondence ABC * — i
ence? How do AC and A'C compare in this case?
B'
BCBC 7 and
A'BC a congru
A'
c
Figure 615 A
On the basis of the preceding discussion, it would appear that if
two sides of one triangle are congruent, respectively, to two sides of a
second triangle and if the corresponding included angles are not con
gruent v then the sides opposite these included angles are not congru
ent and the side opposite the larger angle is the larger side.
It may help you to understand this last statement by examining a
pair of compasses. Observe that, as the angle formed by the pointers
of the compasses gets larger, BO does the distance between the ends of
the pointers. We make this idea formal with our next theorem.
THEOREM 6.11 (SideComparison Theorem for Tivo Triangjtes)
If two sides of one triangle are congruent, respectively , to two sides
of a second triangle, and if the angle included by the sides of the
first triangle is greater than the angle included by the sides of the
second triangle, dieu the third side of the first triangle is greater
the third side of the second triangle.
6,4 Inequalities Involving Triangles 269
restatement: Given A ABC and AA'B 'C w ith AB — A*ff and
BCsrFC. If Z B> LB\ then aC>~FU. (See Figure 616.)
B Hr
C Figure 616
Proof: By the Angle Construction Theorem, there is a point (J> on the
*— >
Oside of AB such that ZAB(<) = ZA'JJ'C", (See Figure 617.) Since
A ABC > ZA'FC, P is in the interior of /.ABC, Choose point P on
such that BF  There are now three cases to consider.
Cose J. P is in the exterior of A ABC,
Case 2, P is on AC
Qwf' 3. F is in the interior of AABC,
We consider only Case 1 here The proofs of Cases 2 arid 3 are left
as exercises.
Proof of Case h In Figure 617, AABP s A A 'B'U by S.A.S. Let BR
he the midray of Z CBP intersecting AC at £. Then APBE ss ACB£
by S.A.S. Applying the Triangle Inequality Theorem to AAEP, we
have AE + EF> AP.
Figure 617
But AP a A'C (Why?) and £P = EC. Why? By the Substitution
Property, we get AE + EC > A'C, and by the Distance Betweenness
Postulate, AC > A'C. Therefore T^C > /TTT and this completes the
proof for Case 1.
270 Inequalities in Triangles Chapter 6
The converse of Theorem 6.1 1 is also true and we state it as our last
theorem of this section.
THEOREM §.12 (AngloComparison Theorem for Two Triangles)
If two sides of one triangle are congruent, respectively, to two sides
of a second triangle, and if the third side of the first triangle is
greater than the third side of the second triangle, then the angle in
cluded by the two sides of the first triangle is greater than the angle
i iK hided by the two sides of the second triangle,
restatement: Given A ABC and AA'B'C with A~B ^ A^W and
BC^&C, if ZC>A 7 &> then /_B> IB . (See Figure
618.)
Figure &I8
Proof: Since m Z B an d nt Z B' are numbers, one of the following must
hold:
(1) mlB<mlB'
(2) mlB=mlB'
(3) ml.B>mLW
Which property of numbers are we using here?
The method of proof is to show that (I) and (2) are impossible, so (3)
must hold* and the theorem will be proved. The problem of showing
that (!) and (2) are impossible is left as an exercise.
EXERCISES 6.4
t. Given A ABC with AZ? = 12, BC = 15, and AC  10. name the. angles
in order of size beginning with the angle of least measure.
2, In A $KM S mlS = 47, mlK = 85 t andml M = 48. Name the short
est side; the longest side.
a Name the longest side of A ABC if (a) ml A = 44, mlB = 90;
(b) ml A = 120, mlB = 40.
4, Does a triangle exist with the following numbers as side lengths? Why?
(a) Si 3, 10 (b) 5, 3, 8 (ci 5, 3, 4
5. Given A ABC with ABD, if ml ABC > mlCBIX prove AC > BC
6.4 Inequalities Involving Triangles 271
6. Given the following figure with angle measures as marked, for each of
the three triangles name the sides of the t dangle in order of increasing
length
7. In the figure for Exercise 6, which segment is the shortest?
8. Given the following figure with angle measures as marked, prove that
CD is the longest segment.
9. In the figure, if the angles have the indicated measures, which segment
is shortest?
In Exercises 1012, got the "best" answer you can in the sense that the
smaller number is as large as possible and the larger number is as small as
possible,
10. Copy and complete: If the lengths of two sides of a triangle arc 7 and
12, then the length of the third side must be greater than Q] and less
than [?].
11. Copy and complete: If the lengths of two sides of a triangle are 6 and 9,
then the third side must have a length less than JTJ and greater than [T,
12. Between what two numbers must the length of the third side of a tri
angle lie if two of its sides have lengths of 17 and 28?
272 Irwqualtties in Triangles
Chapter 6
13, Given ACD, ECB, and with angle measures as marked, prove that
HE > AD.
14. Prove that the sum of the lengths of the diagonals of a convex quadri
lateral is less than the perimeter of the quadrilateral
15, Given AABC with D a
point on AC such that HD
bisects /.ABC, prove that
AB > AD.
16, In the figure, FS < SA
and PQ < QR. Prove that
m£SPQ>m£SRQ.
17, Prove Corollary 6,9.1. (Hint: You will need to use Corollary 6.6.1 and
Theorem 6.9 in your proof.)
18, Prove Corollary 6.9.2. (See Figure 6.9.)
Given AF > PB, prove
m L AC? > m L RCF. {Hint:
Use Theorem 6.12.)
Given AFQR with M
die midpoint of TQ, if
mlRMQ>m/ PMR,
prove that @B > ER,
6,4 Inequalities Involving Triangles
D
'C
273
21. Given a convex quadrilateral ABCD
with AD = BC, if AB > DC com
pare m Z CAD with m Z BCA.
22. Prove the following theorem.
THEOREM TIa Y b,c are the side lengths of a triangle and if h a is the
altitude corresponding to base a, then h a < h and /i a < c.
23. chai.i.fnce problem. Prove Case 2 of Theorem 6.11. This is the Case
whore point P is on SC, (Hint: In Figure fii7 for Theorem 8.11, recall
that AABP=z A A' JVC You must show that AC > A'C.1
24. challenge problem. Prove Case 3 of Theorem 6.11. Tins is the Case
where point P is in the interior of A ABC. (Hint: Let BR he the bisector
ray of ACBP, intersecting AC at E. Why is Ai^E a ACBE? Why is
A£ + BC > A/»? Prove /££ > 5^?.)
25. challenge problem* Complete the proof of Theorem 6. 12 by show
i tig tliat Cases 1 and 2 are impossible. (Hint: Use Theorem 6. 1 1 for Case
1 and the S.A.S. Postulate for Case I.]
CHALLENGE PROBLEM. If P is any
point in the interior of A ABC,, prove
that AP + PB<AC + CB. (Hint:
Let AP intersect W at D. Apply the
Triangle Inequality Theorem to
AACD and to ABDF,)
274 Inequalities in Triangles Chapter 6
CHAPTER SUMMARY
In this chapter we deed I with geometric inequalities involving angles
and sides for any one triangle and also for two triangles. We defined
GREATER THAN and LESS THAN for angles in terms of their measures
and for segments in terms of their lengths.
We stated some order properties for numbers which are listed below
by name only. You should lenow the complete statement of each of these
properties.
01 POSITIVE CLOSURE PROPERTY
62 TRICHOTOMY PROPERTY
03 TRANSITIVE PROPERTY
04 ADDITION PROPERTY
05 MULTIPLICATION PROPERTY
The key theorem concerning geometric inequalities is the EXTERIOR
ANGLE THEOREM which states that an exterior angle of a triangle is
greater than either of its two nonadjaeent interior angles.
Other important theorems involving inequalities in any one triangle
arc listed below by name only, ft is important that you be able to state these
theorems in your own words and that you understand their proofs.
THEOREM 6.8 (77u? Angle Comparison Theorem)
THEOREM 6,9 (The SideComparison Theorem)
THEOREM 6.10 (Tlie Triangle Inequality Theorem)
Tile following two theorems give inequalities concerning two triangles.
Be sure that you know the complete statement of these theorems.
THEOREM 6.11 (The SideComparison Theorem for Two Triangles)
THEOREM 6T2 (27k? AngleComparison Theorem for Two
Triangles)
We defined the DISTANCE between a point and a line not containing
the point to be the long til of the perpendicular segment joining the point to
the line. The distance between a line and a point on the line is defined to be
zero.
We proved that the hypotenuse of a right triangle is the longest side of
the triangle and that the shortest segment from a point not on a line to the
Hue is the segment perpendicular to the line.
Rovltw Exercises
275
REVIEW EXERCISES
In Exercises 17. identify the order property that Is illustrated,
1. If m/ A>mZBandmZ.B = mZC, theu mLA>m£C.
2. If x  y > 5 and y = 4 t then x > &
3. If m£A> mlB, then JtwZA > »»Zfl.
4. If i / y, then % > y or * < y,
5. If A£ < ES and BC < S£ then AB + BC < RS + ST,
6. If i + i/ > z, then x > s — r/.
7. If o + fe > c and a * /> < d t then d > c.
8. In the figure, what must he true about tnAD? About mZ /JF£? AbouL
I
•::
2)
J?
Refer to Figure 619 for Exercises 914. Copy each exercise and replace the
question marks by the symbol < f =, or > to make a true statement.
9, h Q] tl
10. ag]cT]e
11. cQd
12. c[T/
13. fl?J/
14. b^]d[T\f\2\c\2XQ
15. Which theorem do the markings
on the figure contradict?
Figure 619
16. In the figure, which angle is
the largest? The smallest?
276 Inequalities in Triangles
17, In the figure, which side is the longest? The shortest?
Chapter 6
18. In the figure, if the angles have the indicated measures, which segment
is shortest? What theorem did you use to decide?
19, In the figure, if the ingjtea have the indicated measures, arrange the seg
ments PQ, QR, rT5, ?7S, PS, undTTi in order of increasing length.
20, Copy and complete th« fi blowing statement, putting the largest number
in (a) and the smallest number in (b) which will make a true statement
If the lengths of two sides of a triangle are 8 and 15. then the
length of the third side must be greater than (a) (?] and less
than (b) [T]
Review Exercises 277
Exercises 2126 refer to Figure 620 showing A ABC with ADC. BEQ
BFD, and AFE, You should be able to defend your answers using the
orems that you know.
Figure 620
21. Name five angles whose measures arc less than m L 10.
22. Name two angles whose measures are less than mZ.5.
23. Name four angles whose measures are greater than mL 1.
24. L 7 is an exterior uugje of which two triangles?
25. Is /_ 6 an exterior angle of ABFE?
26. Is ml& > mZ2? Explain why.
27. In the figure, if BC is the longest side and 7%\ is the shortest side, prove
that m/,A > m£C, (Hint: Draw GA and use the AngleComparison
Theorem.)
28. Given: SK = KM
ftere KM > KJ
[Hint; Use the Kxterior Angle Theorem, the Isosceles Triangle Theorem,
and the SideComparison Theorem.)
278 Inequalities in Triangles
29. For A ABC prove that AH  WO < AC,
Chapter 6
30. If, In AAjBC, BC = AC, prove that AC > JAB.
31. Given th e fig ure with angle and side measures as marked, name the
sides SR t RQ t and QPin order of increasing length. State a theorem that
justifies your conclusions.
32. Given the figure with sides us marked, copy and insert < or > in (a),
(l>), and (c) so that each is a true statement.
(a) roZAJTJmZBEC
(b) mZDEC[?]mZBEC
(c) ittZJi 0mZDEC
33. Slate theorems that justify your conclusions in Exercise 32,
R*vl*w exercises 279
34. Given: &ABC with median AM
mLAMB = 70
Prove; mABymlC
CMS
35, If ASOf is a right triangle with SMC, prove that KC > KM
K
36. challenge problem. If A, B, C, D are four distinct points in space
and if no three of these points lie on a line, prove that
AB + EC + CD> DA.
37. challenge PROBLEM. Prove that in any triangle there are two sides of
lengths r and s such that
±<t<2.
Dat;kl Phnctltm/Phuto Ke$eurclwr,\
Parallelism
7.1 INTRODUCTION
We are now ready to consider one of the most basic ideas in geom
etry, the idea of parallel lines. What do yon think of when someone says
"parallel lines?" Perhaps the lines which separate lanes on a straight
running track or the cracks between the boards in a floor, or the strings
on a violin? What are some of the properties of parallel lines?
If two ships are sailing parallel courses close to one another as sug
gested in Figure 71 , what is the relationship of the angles marked in
the figure?
Figure 71
282 Parallelism Chapter 7
The congruence of these angles is an example of an Idea from in
formal geometry that suggests a property of parallel lines in formal
geometry.
The set of rails on a railroad track must fit wheels which arc fixed so
that the distance between them cannot change as they roll down the
track. This suggests another property of parallel lines in formal georn
eLry, the property of being the same distance apart everywhere.
What arc the basic properties of parallel lines? Can you think of one
or more properties such that if lines have these properties they should
then be considered to be parallel lines? It would be natural to use such
basic properties in deciding on a definition for parallel lines.
ITie most basic properties of points, lines, and planes are the Inci
dence Properties which were discussed in Chapter L Two different
lines either intersect or they do not, IF they do not intersect and are
copianar, we call them parallel lines. If they do not intersect and are
not copianar, we call them skew lines. We know that two distinct in
tersecting lines determine a plane. It follows that, if two lines are not
copianar, then they must he nonintersecting lines and hence are skew
lines. We shall find it convenient to consider a line as parallel to itself.
We begin our formal treatment with several definitions based on the
foregoing ideas.
7.2 DEFINITIONS
Definition 7.1 Two distinct lines which are copianar and
nonintersecting are parallel lines, and each is said to be paral
lel to the other. Also, a line is parallel to itself. The lines in a
set of lines are said to be parallel lines if each two in the set
are parallel.
Definition 7.2 TWo lines which do not lie in the same plane
are called skew lines.
\ citation. If p and q are lines, then p  q means that p is parallel to (f,
and pX° means that p is not parallel to q.
Note that every pair of distinct parallel lines arc copianar, that
every pair of distinct intersecting lines are copianar, and that every
pair of skew lines are noncoplanar. If p and q are lines, there are four
distinct possibilities:
7.2 Definitions 283
1. p and q are noncoplanar, in which case they axe skt^u .
2. p and q are coplanar but not parallel, in which case they inter
sect in exactly one point and there is exactly one plane which
contains them.
3. p and q arc distinct parallel lines, in which case they do not in
tersect and there is exactly one plane which contains them.
4. p and q are nondistJnct parallel lines, in which case p and q are
the same line and there are infinitely many different planes con
taining p and q t
EXERCISES 7.2
Exercises 15 pertain to the rectangular box suggested in Figure 7*2.
E
Figure 7£
1. Using A, H, C, , . ■ , designate two distinct edges which lie ad parallel
lines.
2. Using A, B, C, . , , , designate two distinct edges which lie on intersect
ing lines.
3. Using A, J3 T C, . , , , designate two distinct edges which lie on skew lines.
4 How many edges of the box lie on lines which are parallel to AB?
5. How many edges of the box lie on lines which are skew to TiB?
6. If p and q are skew lines, why is there no plane which contains both?
7. If p is a line in a plane a t how many different lines in a are parallel top?
8. If p is a line in a plane a t how many different lines in a are perpendicular
top?
0. If p is a line in a plane a, how many different lines in a are skew to p?
10, Prove that two distinct parallel lines detenninc exactly one plane.
11, Given that m is a line and P is a point on tit, prove that there is one and
only one line through P and parallel to RL
12, Let m be a line in a plane a and P a point in a but not on wi. Use your
knowledge regarding the existence of perpendicular lines and the Ex
terior Angle Theorem for triangles to prove that there is at least one
line through P and parallel to m.
284 Parallelism Chapter 7
7.3 EXISTENCE OF PARALLEL LINES
Answering questions regarding existence is an important part of
building a formal geometry. How do we know that there are such
things as poinLs, lines, and planes? We know that they exist because
of the postulates and the theorems of Chapter I on incidence relations.
Given two distinct points A and B on a line l t how do we know that
there is a point Con/ such that ACB and AC  CB? We know it
liecause we can prove it! Our proof depends in a very essential way on
the Ruler Postulate, Given a line A B in a plane a, how do we know that
there exists a ray AC in a such that m/.RAC = 30? We know it be
cause we can prove itl Of course, the proof is easy once we adopt die
Protractor Postulate
Given a line I and a point P not on I, how do we know that there
exists at least one line through P and parallel to J? It is true in some ge
ometries that there are no parallel lines. But in our geometry t the ge
ometry that we inherited from Euclid, there are parallel lines, and
furthermore, we can prove it. In fact, you were asked to prove it in
Exercises 11 and 12 of Exercises 7*2, These exercises arc combined in
the next theorem.
THEOREM 7.1 (Existence of Parallel Lines Theorem) If I is a
line and P is a point, then there is at least one line through P and
parallel to I. If P is on I, Hi ere is exactly one line through P and paral
lel to i
Proof: There are two cases to consider: (1) P is on I and (2) P is not on I
Case 1. We suppose first that P is on if. Since I is parallel to I, it follows
that there is a line through P and parallel to L If m is any line different
from I and through P t then it intersects I in exactly one point and hence
is not parallel to L Therefore there is one and only one line through P
and parallel to I.
Case 2. Suppose next that P is not on I (See Figure 73.) Then P and /
determine a plane. Why? Gall It a.
4 ll
Figure 73
7.3 Existence of Parallel Lines
In a there is a unique line m through P and perpendicular to I (see
Theorem 6.7) and a unique line n through P and perpendicular to m.
We shall prove that n and / are parallel.
Suppose, contrary to what we want to prove, that n and / are not
parallel. Then m, 11, and / are distinct intersecting lines forming a tri
angle with an exterior angle and a nonadjaccnt interior angle both of
which are right angles. (See Figure 74.) But this is impossible in view
of the Exterior Angle Theorem. It follows that n and I arc parallel.
Therefore there is at least one Hue through Panel parallel to I This com
pletes the proof.
285
Figure 74
Along with questions regarding existence in mathematics there are
sometimes questions regarding uniqueness. Theorem 7.1 settles the
matter of existence of parallel lines, but it docs not settle the matter of
uniqueness. If / is a line and .Pis a point on /, we know that there is one
and only one line through F and parallel to L If / is a line and P is a
point nof on I, we do not know yet that there is one and only one line
through P and parallel to L For about 2<XX) years following the time of
Euclid, mathematicians tried to prove that, given a line and a point not
on the line, there is a unique line through the given point and parallel
to the given line. Finally two mathematicians, a Russian named Nikolai
Ivanovitch Lobachevsky (17931856) and a Hungarian named Janos
Bolyai (18021860), proved independently that it is impossible using
only the postulates of Euclid (other than his parallel postulate) to prove
the uniqueness of parallels. Since we want uniqueness of parallels, we
follow hi the footsteps of Euclid and adopt a "parallel postulate," We
defer the statement of lliis postulate, however, to Section 7,8.
In Sections 7.4 and 7.5 we introduce the concept of a transversal
and develop some theorems whose proofs do not depend on the Parallel
Postulate. These arc theorems that belong both to the ordinary geom
etry of Euclid and to the nonEuclidean geometry of Lobachevsky and
Bolyai in which the Euclidean Parallel Postulate is replaced by a postu
late which says that parallels are not unique. More specifically, the
BolyaiLobachevsky Postulate states that if Ms any line and P is any
point not on 1, dien there are at least two distinct lines through P and
parallel to I
28«
Parallelism
Chapter 7
7.4 TRANSVERSALS AND ASSOCIATED ANGLES
Let p, (/, and f be three distinct lines in a plane a. The line t may in
tersect both p and q or it may not If t intersects both p and q, then it
may intersect them in different points or it may intersect them in the
same point In Figure 75, p and q are intersecting lines and t intersects
p and q in different points. In Figure 7G, p and q are parallel lines and
t intersects p and q in different points. In Figure 77, t intersects both
p and q in the same point. In Figure 78, t intersects q but does not
intersect p.
Figure 745
P +
< +
7
Figure 77
Figure 7S
In situations like those in Figures 75 and 76 we say that f is a
trunsc&rsal of p and q. In each of these figures t intersects the union of
p and 9 in a set consisting of two distinct points. On the other hand, in
Figures 77 and 78, t docs not intersect the union of p and q in a set
consisting of two distinct points. As we said, in Figure 75, / is a trans
versal of p and q. In Figure 75 q is a transversal of p and f, and p is a
transversal of q and f. We arc now ready for the following formal
definition.
7.4 Transversals and Associated Angles 287
Definition 7.3 A transversal of two distinct coplanar lines is
a line which intersects their union in exactly two distinct
points.
In Figure 79, A s B t C, D are four of the vertices of a rectangular
box.
li Figure 79
Note that
AB, BC, CD are three distinct lines,
BC intersects the union of .4 B and CD in two distinct
points. Name them.
is not a transversal of AB and CD. Why not?
Is it true that if a line is a transversal of two other lines, then the
three lines arc distinct eoplanar lines? Give a reason for your answer,
A transversal of two lines forms with these lines eight distinct an
gles. For convenience we gjve names to certain pairs of these angles.
Figure 710
In Figure 710, * is a transversal of p and c/; angles 1, 2 t 3, 4 are the
angles formed by p and i; angles 5, 6, 7, 8 are the angles formed by q
and t. Angles 1, 4, 6, 7 are called ulterior angles and angles 2, 3, 5, 8
are called exterior angles ,
2*K
taniliiin
er7
Angles 1 and 6 are one pair of consecutive interior angles. Figure
71 1 shows these angles with several points labeled. Notice that Z 1 is
the union of BA and ED and that Z 1 is LABD. Express Z 6 in terms
of rays and write a name for it involving names of points.
Figure 711 9
Notice that Z 1 and Z 6 have a segment in common and that their
interiors are on the same side of the transversal. Name the segment that
is die intersection of these two angles.
Lines p and q t which we are considering in connection with a trans
versal f, might be parallel or they might not be. if p and q are not paral
lel, then they intersect in a point. This point might be on the opposite
side of the transversal from the interiors of a pair of consecutive in
terior angles, as it would be for Z 1 and Z 6 in Figure 7*11, or it might
be on the same side, as it is for the pair of consecutive interior angles
4 and 7 in Figure 712. Note that although the intersection of Z 1 and
Z 6 is a segment, the intersection of Z4 and Z 7 is the union of a seg
ment ami a set consisting of a single point. Name that segment and that
point. Thus the intersection of two consecutive interior angles may
be a segment or it may be the union of a segment and a set consisting
of a single point.
Figure 712
We shall give a formal definition of consecutive interior angles after
we have introduced two other phrases for angles associated with u pair
of Uiies and a transversal.
7.4 Transversals and Associated Angles 289
In Figure 713, Z 1 and L 7 are interior angles but not consecutive
interior angles and not adjacent angles. We call them alternate interior
angles. The intersection of Z 1 and Z 7 is a segment and their interiors
are on opposite sides of the transversal.
Figure 713
Another pair of alternate interior angles are Z 4 and Z 6, (See Fig
ure 714.) Their intersection is a segment and their interiors lie on op
posite sides of the transversal.
Figure 714
In Figure 715, Z I and Z 6 are consecutive interior angles. Angles
1 and 2 are adjacent angles. Angles J. and 7 are alternate interior angles.
This brings us to Z 1 and £5.
Figure 715
290
Parallelism
Chapter 7
Angle 1 is an interior angle and Z 5 is not. Their interiors lie on the
same side of the transversa]. In Figure 716, these angles are shown
with several points labeled. Examine the intersection of Z 1 and £5.
Figure 7ia
Is the intersection of these angles a ray s the ray BP? Observe Lhat the
intersection of die interiors of Z 1 and Z 5 is the interior of Z 5. Angles
1 and 5 are called corresponding angles.
Another pair of corresponding angles associated with the lines p
and q and their transversal t are Z 4 and Z 8 as shown in Figure 717.
The intersection of these angles is the union of a ray and a set consist
ing of a single point. The interiors of these angles lie on the same side
of the transversal. Although neither interior contains the other, they do
intersect.
Figure 717
Thus far in this section we have introduced alternate interior an
gles, consecutive interior angles, and corresponding angles in connec
tion with two coplanar lines and a transversal. Actually, it is not neces
sary to refer to these lines and the transversal in describing these angles.
Indeed, no such reference is made in the following definitions.
The phrase "a segment and a point*' appears in the following defini
tions. We accept these phrases as a short way of saying "die union of
two sets of points, one of them a segment and the other a set which
consists of a single point/' We accept "a my and a point** to mean "the
union of two sets, one of them a ray and the other a set consisting of a
single point."
7.4 Transversals and Associated Angles 291
Definition 7.4 Two coplanar angles are alternate interior
angles if their intersection is a segment and if their interiors
do not intersect.
Definition 7.5 Two coplanar angles are consecutive interior
angles if their intersection is a segment, or a segment and a
point, and if their interiors intersect.
Definition 7.6 Two coplanar angles are corresponding an
sxlcs if their intersection is a ray, or a ray and a point, and if
their interiors intersect.
EXERCISES 7,1
1, How do yon know that parallel lines exist in Euclidean geometry? Is
your reason a postulate or a theorem or u definition?
& If p is a line, if q is a line, and if p is parallel to q, is it possible that the
intersection of p and q is a set which contains more than one point?
Explain.
3. If p is a line, if q f§ a line, and if the intersection of p and q is the null
set, is it possible that p and q are not parallel? Explain.
4. If p is a line, if q is u line, and if p is parallel to q, is it possible that there
is no plane containing the union of p and q? Explain,
■ Figure 718 sh o ws t wo coplanar lines p a nd q a nd a transversal t with several
points labeled. Copy Exercises 5 10 and replace the question marks with
one word,, two words, or three capital letters so that the resulting statement
is a true sentence concerning Kijjyre 718.
Figure
5* LBDC and ZQ arc corresponding angles,
fl, / BDC and Z[7] are alternate interior angles.
7. L BDC and Z T arc consecutive interior angles,
a L FDC and LAQD are \?J angles.
9. / HDF and Z HOP. are CD angles.
10, Z FDC and / ECD arc [T angles.
292
Parallelism
Chapter 7
Figure 7 19 shows two copbnur lines p and q and a transversal t. Eight an
gles are labeled. Copy Exercises 1 1 20 and replace the question marks with
one word or two words or a letter or a number so as to make a true statement
about Figure 719,
BrgHK 719
11. a and C are vertical [?].
1& a and d are \T\ angles.
13. a and u are [?] angles,
14. c and c are [7 angles.
15. c and [7] are corresponding angles.
16\ x and [f are corresponding angles,
17. d and [?] are alternate interior angles.
18. There are JT pairs of alternate interior angles among the eight angles.
19. Thereare {7] pairs of consecutive interior angles among the eight angles.
20. There are \T\ pairs of corresponding angles among the eight angles.
<■ — ■» < — * «. — t
Figiue 720 shows three iioncollineur points A, B f C, the lines AB, HC, CA,
and 12 angles formed by than, Copy Exercises 2123 and replace the ques
tion marks so as to make a true .statement about the figure.
Figure 720
21. / 1 and ]7] are corresponding angles; also / 1 and [Tj are corresponding
angles.
22. Z 12 and \T\ are alternate i ntcrior angles; also L 12 and [?] are alternate
interior angles.
7,4 Transversals and Associated Angles 293
23. _ 1 2 and IT are consecutive interior angles; also / 12 aud [T] are con
secutive interior angles,
Figure 721 shows two distinct parallel lines, a transversal, and eight associ
ated angles. Copy Exercises 2428 and replace the question marks so as to
make a true statement about the figure.
"4
Figure 721
24, Angles 1 and 5 arc corresponding angles; their intersection is Q]; I heir
interiors lie on [7] of L
25, Angles 3 and 5 are consecutive interior angles; their intersection is [Tj:
their interiors lie on [7] of t,
26, Angles 3 and 6 are alternate interior angles; their intersection is Q]; their
interiors lie on [7] of f ,
27* Angles 6 and 7 are vertical angles; their union is the [tj of a pair of lines;
their intersection is Q]; the intersection of their interiors is [7],
28. Angles 5 and ft are a linear pair of angles; their union is the union of a
line and a [T; their intersection is JT]; the union of these angles and their
interiors is the union of a halfplane and Q],
20, The figure shows A ABC and two rays in the same plane such that B is
between A and E r and BD is between liE and EC. Considering only
angles that can be named in terms of three points labeled in the
figure, identify all pairs of alternate interior angles,
30. Under the same conditions as those in Exercise 29„ identify ail pairs of
consecutive interior angles,
31. Under the same conditions as those in Kxercise 29, identify all pairs of
corresponding angles.
294 ParalJefism
Chapter 7
Figure 722 shows a quadrilateral and one of its diagonals. In Exercises
3235, copy and complete each statement so that it will be true.
32. / ACH and Z CAD are alternate interior angles detennmed by trans
versal aS and lines \f\ and [7].
33. Z BAC and / DC A are alternate interior angles determined by trans
versal TJ and lines \T\ and {T},
34 Z ABC and Z BCD arc \T} angles determined bv transversal jT] and lines
(U and Q] .
35. Z BAD and Z arc consecutive interior angles detennined by trans
versal [T\ and lines and [7).
Figure 723 is a plane figure with five angles labeled, including two pairs
of vertical angles. Let m Z a m 65 and mZa = m / e. In Exercises 3638 t
find the measure of the given angle
36. Li
37. Zu
38. Ze
Figure 723
Figure 724 is a plane figure showing two lines and a transversal. Refer to
tills figure for Exercises 3942.
Figure 724
7.5 Some Parallel Line Theor&m*, 295
39. Name
(a) the pairs of alternate interior angles;
(b) tlie pairs of consecutive interior angles;
(c) the pairs of corresponding angles.
40. Prove: If one pair of alternate interior angles are congruent, then
(a) the other pair of alternate interior angles are congruent;
(b) each pair of consecutive interior angles are supplementary:
(c) each pair of corresponding angles arc congruent
41. Prove: If one pair of consecutive interior angles are supple mentary,
then
(a) the other pair of consecutive interior angles are supplementary;
(b) each pair of alternate interior angles arc congruent;
(c) each pair of corresponding angles are eongruent
42. Prove: If one pair of corresponding angles are congruent, then
(a) each pair of corresponding angles are congruent;
(b) each pair of alternate interior angles arc congruent;
(c) each pair of consecutive interior angles are supplementary,
7.5 SOME PARALLEL LINE THEOREMS
This section contains several theorems that are sometimes helpful
in proving that two lines are parallel.
THEOREM 7,2 Let two distinct coplanar lines and a transversal
be given. If the transversal is perpendicular to both lines, then the
lines are parallel,
Proof: Let a and b be distinct coplanar lines and let t be a transversal
of them. (See Figure 725.) We wish to prove that if a and h are per
pendicular to t t then a and b are parallel. Suppose, contrary to this as
sertion and as suggested by the figure, that a _L t. that feif, and that
a is not parallel to b. Then a and b intersect in some point P forming
APVQ. This triangle has Z 1 as an exterior angle and Z 2 as a nonadja
cent interior angle. By hypothesis, both of these angles arc right angles,
so &F¥Q is a triangle with one exterior angle congruent to a nonadja
cent interior angle.
296 Parallelism Chapter 7
According to the Exterior Angle Theorem, however, the measure
of an exterior angle of a triangle is greater than the measure of cither
nonadjacent interior angle 1 truce we have two angles which are con
gruent by one line of reasoning and which are not congruent by an
other line of reasoning. We arrived at this contradiction after we had
supposed that die lines a and h are not parallel Since we cannot have a
contradiction in our system, we have proved that lines a and h cannot
lie not parallel. Hence they are parallel This completes the proof of
the theorem.
We pause briefly to comment on this proof. It is an example of an
indirect proof, The theorems in our formal geometry arc true state
ments. Their truth rests on a foundation of postulates and definitions.
In proving theorems, that is, in establishing their truth, we may use
definitions, postulates, and theorems proved previously.
flow did we prove Theorem 7i2? Consider the given situation in
volving lines a T h r and t. Lines a and h are distinct and coplanar, and /
Is perpendicular to both of them. We do not know that a and h are par
allel nor do we know that a and b arc not parallel But we do know
from our definitions that a and b are either parallel or they are not par
allel Since one of these possibilities leads to a contradiction, we con
clude that the other possibility is the valid conclusion.
It may be helpful to see this reasoning in skeleton form using sym
bols. We start with a situation in which a and b are coplanar lines and t
is a transversal of them. Let P and Q stand for statements as follows:
P: t ±a and t 1 b.
Q. tffc
Then our theorem and proof arc essentially as follows.
THEOREM If P, then 0.
Proof: It is given that P is true. Either Q is true or Q is false. If QU
false, then it follows that Ph false, This contradicts the hypothesis that
P is true, Therefore Q is true, and the proof is complete.
If P denotes a statement, it is convenient to denote the opposite
statement by notH If P is a true statement, then notF is a false state
ment If P is a false statement, then notP is a true statement. If a state
ment has the form "If P, then Q" then there is always a related state
ment of the forrn "If notQ* men notJV' This related statement is
called the contrapositivc of the given statement. If the contrapositive
of a statement is true, then the statement is true. For, if notQ implies
notP, then the only way it is possible for J* to be true is for {) also to be
7.5 Some Parallel Line Theorems 297
true. In other words, if notQ implies notP, then P implies Q. This
means that we can prove a theorem by proving its contrapositive. If it
is easier to prove the eontrapositive, then this is what we should do. A
proof using the contrapositive is one form of indirect proof.
For example, suppose we wish to prove
if a 2 =f=b 2 s then a^=h.
An easy way to prove this is to prove its conlrapositive
if a = h, then a 2 = h 2 .
Tlie proofs of the following three theorems are assigned as exercises.
THEOREM 7,3 {Alternate Interior Angle Theorem) If two al
ternate interior angles determined by two distinct coplanar lines
and a transversal are congruent then the lines are parallel.
THEOREM 7.4 (Corresponding Angle Theorem) If two corre
sponding angles determined by two distinct coplanar lines and a
transversal are congruent, men the lines are parallel.
THEOREM 7.5 [Comecutwe Interior Angle Theorem) If two
consecutive interior angles determined by two distinct coplanar
lines and a transversal are supplementary, then the lines are parallel
EXERCTSES 7.5
Figure 726 shows u transversal of Iwo distinct coplanar lines and eight as
sociated angles, In Exercises 1—4, measures of two of the angles are given.
Explain why m must lie parallel to n.
m +
««
Figure 726
1. m/.a = 105, m£e = 105 3. mLc = 75, mLv = 105
2. mid = 105, mZf= 105 4. m t a = 75, m L f = 75
29S
Parallelism
Chapter 7
Figure 727 shows five coplanar segments and several associated angles. In
Exercises 59, state which lines must be parallel on the basis of the given
information.
5. Z«s Iv
6. Zpss lq
7. ml ADC + mlDCB = 180
8. ml ADC + ml BAD = ISO
9. ml DAB + ml ABC = 180
Figure 728 shows five distinct coplanar points A, B, Q D, E. Points A, B, C
are coHinear, AD 1 Tkff, and CE _L AC In Exercises 1014, state whether
this given information implies the Stated conclusion.
10. A3  AC
11. A~$  M
12. a6 if ci
13. AD  EC
14. XBl\ BE
In Exercises 1520, the figures show three coplanar lines. The measures of
three angles, expressed in terms of a number x, are given in the figures Find
x and determine whether m is parallel to n. Give a reason for your answer.
15.
16.
7,5 Some Parallel Una Thtortms 299
18.
m4
"4
2x + 5
™«
lOO + i/ 80 + *
*4
100 + 21
21. Given: A plane figure with AB = CD, AD = BC.
Trove: AD \ B&. XS jj DC
SOD
Parallelism
Chapter 7
22. Given: A figure with ADB, AEQ
AD = A£ AB = AC.
LACBz* IADE
Prove: DE\BC
23. Given: A figure in which A~C and SD bisect each other at M,
Prove: A~B \ S3, A/5  S3
2? C
24. Gitcn. A figure with ARB, BPC, CQA.
AQ QC= RP
CF = PBzzQR
RR = RA = PQ
Prove: m/_A + m£B + m I C = 180
25. Git en. A plane figure, with I A =z IB, AD = BC, A£ = EB t
DF=m
Prove: EF 1 W, AB J EF, DC  A~S
U Lil £ III .C
A £ rt
26. Prove Theorem 7.3.
27. Prove Theorem 7,4.
28. Prove Theorem 7.5.
29. Write the contraposiuve of Theorem 7,3. Is this coiitrapositive a true
statement?
30. Write the contrapositive of Theorem 7.4. Is this contrapositive a true
statement?
31. Write the contrapositive of Theorem 7.5. Is this contrapositive a true
statement?
7.6 The Parallel Postulate 301
7.6 THE PARALLEL POSTULATE AND SOME THEOREMS
Theorem 7. 1 states that through a given point not on a given line
there is at least one line parallel to the given line. As we pointed out
following the proof of that theorem, it is impossible, using only Pos
tulates 125, to prove that this parallel line is unique. Since we
want it to be unique, we adopt the following Parallel Postulate. As we
stated in Section 7.3 it is this postulate which makes our geometry,
Euclidean geometry, different from that of Lobachevsky and Bolyai.
POSTULATE 26 (Parallel Postulate) There is at most one line
parallel to a gjven line and containing a given point not on the given
line.
This postulate and Theorem 7.1 tell us that if a line and a point not
on it are given t then there is exactly one line through the given point
and parallel to the given line. Furthermore, as we said in Section 7.3,
we know that if a line and a point on it are given, then there is exactly
one line through the given point and parallel to the given line, namely
the given line itself. Hence, in every case, given a line and a point,
there is exactly one line through the given point and parallel to the
given line*
In Section 7.5 there are some parallel line theorems, actually the
orems stating conditions that imply that lines arc parallel. These the
orems are useful in proving lines parallel. In this section we have the
converses of several of these theorems, These converses were deferred
until now because the Parallel Postulate is essential for their proof.
If a theorem is of the form "If P, then Q," then its converse is the
associated statement "If Q, then ?" It should be clear that the con
verses of some true statements are not true statements. For example,
'Tf a number is greater than 100, then it is greater than 6" is a true
statement, whereas its converse, c *If a number is greater than 6, then it
is greater than 100," is certainly a false statement The converses of
some theorems are theorems: the converses of other theorems are not
theorems. We shall prove that the converses of Theorems 7.3, 7.4, and
7.5 are theorems.
Note first, however, that the converse of Theorem 7.2 is not true.
To see this, suppose m and n are distinct parallel lines and that f is a
transversal of them which makes an augle of measure 30 with m. Then,
according to Theorem 7.7 given later in this section, it also makes an
angle of measure 30 with line n. Thus wc see that lines m and n may be
parallel even though a transversal is not perpendicular to both of them.
We proceed now to the converses of the other theorems in Section
7.5.
302 Parallelism Chapter 7
THEOREM 7.6 (Converse of Alternate Interior Angle Tlieorem)
Tf two distinct lines are parallel then any two alternate interior
angles determined by a transversal of the lines are congruent.
Proof: (See Figure 729.) Let m and n be distinct parallel lines. Let t
be a transversal that intersects them in P and Q, respectively. Let R
and S be points on m and n, respectively, such that R and S are on op
posite sides of t. Wc shall prove that LRPQ == LSQT.
t
Figure 7W
It follows from the Angle Construction Theorem that there is a ray
~PR\ with W on the tfside of U such that LR'PQ as Z 5QK Since
Z RTQ and Z SpP are congruent alternate interior angles, it follows
from Theorem 7.3 that lines AT and n are parallel. Therefore AT and
m are lines through P and parallel to n. From the Parallel Postulate,
however, it follows that there is only one line through F and parallel
to n. Therefore iiT and m are the same line. Then Z R'PQ and Z RPQ
are the same angle and it follows that Z RPQ m Z SQP.
THEOREM 7, 7 (Converse of Corresponding Angle Theorem) Tf
two distinct lines arc parallel, then any two corresponding angles
determined by a transversal of the hues are congruent.
Proof; Assigned as an exercise.
THEOREM 7,8 (Converse of Consecutive Interior Angle The
orem) If two distinct lines are parallel, then any two consecutive
interior angles determined by a transversal of die lines are
supplementary.
Proof: Assigned as an exercise.
As we noted above, the converse of Theorem 7.2 is not true. It
should be noted, however, that a theorem similar to Theorem 7.2 does
have a true converse. This is Theorem 7.9. Its converse is Theorem
7.10. Tlieorem 7.9 follows immediately from Theorem 73. Theorem
7.6 The Parallel Postulate 303
7.2 begins with a "situation statement" (Let two distinct coplanar lines
and a transversal he given) followed by an "If P, then Q'* type of .state
ment In Theorem 7.9 some of the P has been put into the "situation/'
but the meaning of the two sentences taken logcdier is the same as in
Theorem 7.2. Theorem 7.10 follows immediately from Theorem 7.7.
THEOREM 7.9 I^et a and h be two distinct coplanar lines, and
let t be a transversal of them that is perpendicular to a. If t is per
pendicular to b, the lines a and b are parallel.
THEOREM 7.10 Let a and h l>e two distinct coplanar lines, and
let f he a transversal of them that is perpendicular to a. If a and b
are parallel, then f is perpendicular to h.
The next theorem provides anodicr useful method for proving lines
parallel,
TTTEOR EM 7,11 Two coplanar lines parallel to the same line are
parallel to each other.
Proof: Let p and q be coplanar lines each parallel to a line r. If p — q t
then p is certainly parallel to q. Suppose, then, that p and q are distinct
lines. There are two possibilities as indicated in Figure 730. Either p
and q are parallel lines or else they have exactly one point T say P, in
14
P'oeelbUity (») Powubility fb)
Figure 730
common. If they have exactly one point P in common, then there arc
two distinct lines through P and parallel to r. Since this contradicts the
Parallel Postulate, it follows that possibility (h) in Figure 730 is impos
sible, and therefore p is parallel to q.
If we consider only lines lying in a given plane, we see that the re
lationship of parallelism for lines is an equivalence relation. That is.. It
is reflexive, symmetric, and transitive, (1) It is reflexive since every line
is parallel to itself. (2) It is symmetric since if line p is parallel to line </,
then line q is parallel to line p, (3) To show that it is transitive, suppose
that line p is parallel to line r and that line r is parallel to line q. Then,
since parallelism in a plane is symmetric, it follows that line q is par
allel to hue r. Then p and q are both parallel to line r and it follows from
304
Parallelism
Chapter 7
Theorem 7,11 that p is parallel to q. Later we shall see that parallelism
for lines is an equivalence relation even without the restriction that the
lines all lie in one plane.
THEOREM 7A2 Let three distinct eoplanar lines with two of
them parallel be given. If the third line intersects one of the two
parallel lines, then it intersects the other also,
Proof: Assigned as an exercise.
THEOREM 7.13 Let two sets % and 3 of parallel lines in a plane
q he given, (This means that every two lines in § are parallel and
that every two lines in 3 are parallel.) If one line in § is perpendic
ular to one line in 3 , then every line in § is perpendicular to every
line in 3.
Proof: Let s be a line in S and let t be a line in 3 such that s is per
pendicular to t. Let u be any line in g and let v be any line in 3 . We
want to prove that u is perpendicular to c. Suppose, first, that s and u
are distinct lines and that r and i" are distinct lines as indicated in Fig
ure 731. Since * intersects t and i is parallel to t% it foDows from The
orem 712 that 8 intersects v and therefore s is a transversal of the par
allel lines t and o. Then it follows from Theorem 7.10 mat s is
perpendicular to v. Similarly, it follows from Theorems 7.12 and 7.10
thai c is a transversal of the parallel lines * and u and that v is perpen
dicular to u. This completes the proof for the case in which s and u are
distinct lines and t and are distinct lines.
i
►
8

u«
►
Figure 731
Suppose next that s = u and that s is perpendicular to t. Then it
follows as above (really the halfway point in the reasoning of the pre
ceding paragraph) that s is perpendicular to v.
Tlie ease in which s and u are distinct and t = c is assigned as an
exercise.
The case in which s = u and I = v requires no proof since the
hypothesis and the conclusion are the same in this instance.
7.6 The Parallel Postulate
305
EXERCISES 7,6
1. Use Theorem 7,6 to prove Theorem 7 J.
2. Use Theorem 7.6 to prove Theorem 7.8.
3. Prove Theorem 7.7 without using Theorem 7.6, (This proof will be sim
ilar to I he one for Theorem 7.6 that uses the Parallel Postulate.)
4 Prove Theorem 7.8. (This proof will be similar to the one for Theorem
7.6 that uses the Parallel Postulate.)
5. Explain (he following statement; Of the three theorems, 7,fi, 7,7, 7.8 t
any one of them may be considered as the basic theorem and the other
two as corollaries of it.
6. Prove Theorem 7.13 for the case in which a =5^ u and t = c.
In Figure 732, lines o, b, c, and d are eoplanar and the measures of several
angles are marked. In Exercises 79, justify the given assertion regarding
the lines in this figure.
Figure 7:12
7, a is parallel to h.
8. a is parallel to c.
9* h is parallel to c.
10. JSa t b,c are eoplanar lines such that a is parallel to b and b is parallel to
c+ does it follow that a is parallel to e? Is this an instance of the reflexive
property of parallelism for lines? Of the symmetric property? Of the
transitive property?
11* If a t b, are eoplanar lines such that a is parallel to b and b is parallel to
c, docs it follow thai b is parallel to n? Is this an instance of the reflexive
property of parallelism for lines? Of the symmetric property? Of the
transitive property?
12. If a, h. c are eoplanar lines such that a is parallel to b and b is parallel to
C does it follow that c is parallel to c? Is tilts an instance of the reflexive
property of parallelism for lines? Of the symmetric property? Of lite
transitive property?
306
Paralii 
Chapter 7
In Figure 733, a, b t c, d are coplanar parallel lines, f is a transversal of ihcm,
and the measures of se%'eral angles are marked. In Exercises 1318, justify
the given assertion regarding angle measure.
13. x= 117
14. y = 63
15. z  117
16. u  63
17. o = 63
18. w = 63
af
b +
t +
d +
J
Figure 73.1
in F.xerciscs 1921, use Figure 734 with EAD, m Z EAB = 1 IS, KD  BC.
19. Find™ lABDHZm LABD = 3m Z.DBC.
20. Find m /DBC and in ZBDA if 2m
lABD = 3m I DBC
22. Given a plane figure made up of two parallel lines, a transversal, and a
segment, with angle measures as marked, find x\ y, and z.
23. Given a plane figure made up of two pairs of parallel lines, with angle
measures as marked, find x, y, z, u, v, and to.
7.6 The Parallel Postulate
307
24. Given a plane figure with AB DE. with A. C, E all on the same side of
BD, with C on the fcside of AB, and with C on the A side of ED, prove
that
m Z BCD = m / ABC + m Z COE
(Hint Copy the Figure and draw the line
through C parallel to 12.)
4 — *■ n
25, In the figure for Exercise 24, what is the sum of the measures of Z ARC,
LCBD, LBDC&ad /LCDE?
26, In the figure for Exercise 24, what is the sum of the measures of Z CUD*
ZBDC. and ZDCB?
27, The figure helow shows some segments that are parallel and some that
are not parallel. Try to write a good definition of parallel segments.
Punt"!,:
Parallel Not pimiltet V Not paralM
25. The figure below shows some quadrilaterals that are parallelograms and
some that are not. Try to write a good definition of parallelogram.
308 Parallelism : ::  ~
7.7 PARALLELISM FOR SEGMENTS; PARALLELOGRAMS
In this section we introduce some definitions and theorems con
cerning an important type of quadrilateral called a parallelogram. Since
the sides of a quadrilateral and, hence, of a parallelogram are segments.,
we need a formal definition that extends the concept of parallelism to
segments. In Section 7,8 we extend the concept also to rays.
Definition 7.1 If the lines which contain two segments are
parallel, then the segments are said to lie parallel segments.
and each is said to be parallel to the other. The segments in a
set of segments are parallel if every two of them are parallel.
The lines which contain parallel segments need not be distinct. In
other words, a segment of a line is parallel to every segment of that
line. As a special case of a special case, we note that every segment is
parallel to itself.
Let A, B, C t D be four points with A^B and C ^ D. {See Figure
735.) Then KB is parallel to TJD if and only if AS is parallel to CD. We
use the same symbol to denote parallelism for segments that we use for
lines. Thus AB . CD means that AB is parallel to CD.
C D
t ^i ► Figure 735
Definition 7.8 A parallelogram is a quadrilateral each of
whose sides is parallel to the side opposite it
Consider a parallelogram ABCD as shown in Figure 736. Since
ABCD is a quadrilateral, it follows that AB, BC, CD, DA are four dis
tinct lines Since An and CD are distinct parallel lines, it follows that
C and D He on the same side of AB* Similarly, D and A lie on the same
side of BC, A and B lie on the same side of CD, and B and C lie on the
same side of DA. Therefore each side of a parallelogram lies on a line
which is the edge of a halfplane that contains all of the parallelogram
D C
Figure 736
7.7 Pararffelitm for Segments; Parattetograrm 309
except that side. Therefore every parallelogram is a convex quadri
lateral. (The word convex is used here in the sense of a convex polygon.
See Definition 4, 1.6.)
The next three theorems state some important properties of
parallelograms.
THEOREM 7.14 If a convex quadrilateral is a parallelogram,
then its opposite sides are eongnient.
Proof: Let parallelogram ABCD
be^ given. (See Figure 737.) Then
AB i CD and BC \\ DA. We shall
prove that AB = CD and that
BC = DA. First draw the seg
ment A€, Our plan is to use con
gment triangles,
Statement
1, ZBACs IDCA
% ACzsZA
3. AACBsa LOAD
4. AAiJCss &CDA
5. ab ^ £E anc i bc ^ DA
Reason
1. Alternate interior angles de
termined by a transversal of
two parallel lines are congruent,
2. Why?
3. Why?
4. Why?
5. Why?
JHEOREM 7.15 If a convex quadrilateral is a parallelogram, then
its opposite angles are congruent.
Proof: Assigned as an exercise.
THEOREM 7, 16 If a convex quadrilateral is a parallelogram, then
its diagonals bisect each other.
Proof: Assigned as an exercise.
The next three theorems are useful in proving that certain quadri
laterals are parallelograms.
THEOREM 7.17 If two sides of a convex quadrilateral are parallel
and congruent, then the quadrilateral is a parallelogram.
Proof: Assigned as an exercise.
310 Parallelism
Chapter 7
TITEORFM 7J8 If the diagonals of a convex quadri lateral bisect
each other, then the quadrilateral is a parallelogram.
Proof: Assigned as an exercise.
THEOREM 7.19 If each two opposite sides of a convex quadri
latend are congruent, then the quadrilateral is a parallelogram.
Proof: Assigned as an exercise.
Just as parallelograms are convex quadrilaterals with a special prop
erty, so are trapezoids convex quadrilaterals with a special prop
erty (only not quite as special as that for parallelograms).
Definition 7.9 A trapezoid is a convex quadrilateral with
least two parallel sides.
Note that we do not say two and only two sides parallel. A trape
zoid may have only one pair of parallel sides or it may have two pairs
of parallel sides. In other words, every parallelogram is a trapezoid, but
not every trapezoid is a parallelogram. In some l*x>ks trapezoids are
restricted to have only two parallel sides. In these instances the set of
all trapezoids and the set of all parallelograms do not intersect. In this
book the set of all parallelograms is a subset of the set of all trapezoids.
Rhombuses, rectangles, and squares are all parallelograms with
special properties. Their formal definitions come next.
Definition 7.10 A rhombus is a parallelogram with two ad
jacent sides congruent
Definition 7.11 A rectangle is a parallelogram with a right
angle.
Definition 7.12 A square is a rectangle with two adjacent
sides congruent.
THEOREM 7.20 A rhombus is an equilateral parallelogram.
Proof Assigned as an exercise.
7,7 ParalltHsm tor Segments; ParallefogTams 311
THEOREM 7.21 A rectangle is a parallelogram with four congru
ent angles.
Proof: Assigned as an exercise.
THEOREM 7.22 A square is an equilateral rectangle.
Proof; Assigned as an exercise,
THEOREM 7,23 A square is an equiangular rhombus.
Proof: Assigned as an exercise.
THEOREM 7.24 The diagonals of a rhombus are perpendicular.
Proof: Assigned as an exercise.
In order to prove that a figure is a rhombus it is sufficient to prove
that it is a parallelogram with two adjacent sides that are congruent
(Definition 7.10). Tf we know that a figure is a rhombus, then we may
conclude that all four of its sides are congruent (Theorem 7.20), If we
want to show that a parallelogram is a rectangle, it is sufficient to show
that it has one right angle, since it then necessarily has four right angles
(Definition 7,11 and Theorem 7,21' lb show that a rect&ngfc is a
square, it is sufficient to show that two of its adjacent sides are congru
ent, since then all four of its sides are congruent (Definition 7.12 and
Theorem 7.22}. To show that a rhombus is a square, it is sufficient to
show that it has a right angle since it is easy to show that an equilateral
parallelogram with a right angle is an equilateral rectangle, that is, a
sqiiare.
In Chapter 3 we introduced the concept of distance. We restricted
ourselves there to the idea of the distance between two points. We are
ready now to extend the idea oi' distance to the distance between two
parallel lines. Of course, we agree that the distance between a line and
itself is zero. So let us consider the idea of the distance between two
distinct parallel lines m and n as suggested in Figure 738. The length
of a segment (sec the "dashed" segment in the figure) joining a point
of one line to a point of the other line might be very long depending
on bow the endpoiuLs are picked.
« i v^ 1 1
312
Parallelism
Chapter 7
It seems natural to think of the distance between two parallel lines
as the length of the shortest segment joining the two lines, and it seems
that this segment should be perpendicular to both lines. If we pick any
point on m, say M, then there is a line t in the plane of m and n that is
perpendicular to m at M , (See Figure 739.) Also I intersects n in some
point* say A 7 , and MN is perpendicular to both m and n. If t' is another
transversal of m and n perpendicular to both of them and intersecting
them in M' and V, respectively, how do we know that M.N and M'N'
are congruent? Our idea for the distance between two parallel lines
will not be any good unless we can show that MN and M'S 7 are con
gruent. The next theorem shows that this is possible.
Figure 739
THEOREM 7.25 For every two distinct parallel lines there is a
number that is the common length of all segments perpendicular to
both of die given lines and with one end point on one of the given
lines and one eiidpotnl on the other one.
Proof: {See Figure 740.) I>et m and n be two distinct parallel lines.
a b
Figure 740
Let A and B be two distinct points of m, and C and D two distinct
points of n such that the segments AC and BD are perpendicular to
both m and n. Then m is a transversal of AC and BD and is perpendicu
lar to both. It follows that KC and WD are parallel (Theorem 7.2); hence
ABCD is a parallelogram (by definition) and AC = BD (Theorem 7.14),
If we think of segment KC in Figure 740 as a fixed segment and BD
m a variable segment (one that we can pick anywhere as long as B is
on m, D is on n, and ¥D is perpendicular to m and to n), then the num
ber we are looking for, the one whose existence wc wanted to prove, is
the number AC. The idea of Theorem 7,25 is sometimes expressed by
saying that two parallel lines arc everywhere equidistant.
We are now ready for the following definition.
7.7 Parallelism for Segments; Parallelograms 313
Definition 7.13 The distance between two distinct parallel
lines is the length of a segment which is perpendicular to both
lines and whose endpoints lie on these lines, one endpoint on
one line and the other endpoint on the other line. The dis
tance between a line and itself is zero.
EXERCISES 7.7
1. Frove: If the convex quadrilateral ABCD is a parallelogram, then Z A fix
Z C and I B as ID. (This is Theorem 7.15.)
2. Prove: If the convex quadrilateral ABCD Is a parallelogram, then A~C
and WD bisect each other. (This is Theorem 7.16.)
3. Frove: II ABCD is a convex quadrilateral, if AB m CH5, and if AB  CD,
then ABCD is a parallelogram. (This is Theorem 7.17.)
4. From: If ABCD is a convex quadrilateral, and if AC and #D bisect each
other, then ABCD is a parallelogram. (This is Theorem 7.18,)
5. Frove: If ABCD is a convex quadrilateral if AB ==? CD and All as BC t
then ABCD is a parallelogram. (This is Theorem 7.19.)
6. If ABCD is a rhombus and if AB =s AD, then AB = BC = CD = DA
and ABCD is an equilateral parallelogram. {This is Theorem 7.20.)
7. Prove Theorem 7.21. 9. Prove Theorem 7.23.
8. Prove Theorem 7.22. 10. Prove Theorem 7.24,
11. Why are opposite sides of a parallelogram parallel?
12. Why are opposite sides of a parallelogram congruent?
13. The figure below shows some rays that are parallel and some that are
not parallel, Try to write a good definition of parallel rays.
A B c • > 4 • • »
AB and BC are parotid Parallel rays Not parallel rays Not parallel nyt
14. The figure below shows some antiparallel rays and some rays that are
not antiparallel. Try to write a good definition of antiparallel rays.
A B CD
— • ► 4 — • — • • m
BA and EC
are antiparalle]
*
?Cot antiparaU*!
BA and CD
are antipartllel
B
ACnndAB
are not aariparalM
4
Antiparallrl my*
Not antiparaCe]
314
Parallelism
Chapter 7
7,8 PARALLELISM FOR RAYS
In this section wo extend the concept of parallelism to rays. Since
segments are parallel if the lines containing them arc parallel, we might
consider rays parallel if the lines containing them are parallel. But it is
useful to Distinguish between the case suggested by rays AB and CD
and the case suggested by rays EF and GH in Figure 741. In both
cases the rays lie on parallel lines. In one case, however, they point in
the same direction, whereas in the other case they point in opposite
directions. (We use the word "direction" here in an intuitive sense; it is
not a part of our formal development*)
Figure 741
Definition 7A4 (Sec Figure 742.) Two noneollincar rays
AB and CD are parallel if AB and CD are parallel lines and
if B and D lie on the same side of AC. Two coDmear rays are
parallel if one of them is a subset of the other.
Definition 7,15 (See Figure 742.) Two noneollincar rays
EF and GH arc anliparallel if EF and GH are parallel lines
and if F and H lie on opposite sides of EG, Two collinear rays
are an ti parallel if neither is a subset of the other.
AB unci CD urv porulkl.
Figure 742
IJ*n±JK
JIutAIKm
iirti pontile].
EF mul GH are antiparalkL
The proofs of tile following three theorems are assigned as ex
ercises.
THEOREM 7.26 If Z A BC and I DEF are coplanar angles with
BA and eB parallel and b8&xuI EF* parallel, then Z ABC^ Z DEF.
7.8 Paraltellsm for Rays 315
THEOREM 7.27 If Z ABC and t DEE are coplanar angles with
— * — * ■ — ^ — *
BA and ED parallel and with BC and Eb antiparallel, then / ABC
and LDT.E are supplementary.
THEOREM 7.2S If / ABC and Z OFF am coplanar angles with
BA and ED antiparallel and with BC and EF antiparallel then
LABC^ A DEE
EXERCISES 7.S
Dra%v a line and on it mark five points A, B, C, D, £ in the order named. In
Exercises 120, slate whether the given .statement is true or fake.
2. AB = 52
3. 31 =s M
4. AB = M
5, AB  BA
6. AB  AC
7, IS  CD
&.2Sc3ft
9. AB c BC
10, a2 C AC
ilaScbS
12. 5? C AB
13, Ai! and BA arc parallel.
14* An and BA are anti parallel.
13, AB and FD are parallel.
16. HC and BA are antiparallel.
17. b? n bI = {B}
18. Bf? f DA = BD
19. BC? P DA = Wi
20. d£ 1CD=
A t B T C, . . . , K, L are distinct coplanar points, as in Figure 743, so that
t j ^ ) <  ^ j
DF  j/i and D/f  Fi. In Exercises 2130, state whether the given rays are
parallel or antiparallel,
21. 15i and 77?
22. DE and 77?
23. £// and BE
24. I1A and Li
25. ^andEF
26. CF and FC
27. CF andlf
28. Cpmidjf
29. Gland 77f
30. GfmidjH
Hfure743
316 Parallelism Chapter 7
■ In Exercises 3136, copy and complete the statements.
31. The intcrseetitm of two parallel collinear rays is [?].
32. The intersection of two parallel noncollinear rays is [7],
33. The intersection of two antiparallel collinear rays is (?] or [Tj or (?].
34. The union of two parallel collinear rays is (t).
35. The union of two anti parallel collinear rays is a line with the interior
points of a segment deleted or it is an entire fT.
36. If ABVD is a parallelogram, then An and CD are [7] rays.
37. Prove Theorem 7.26 for the case in which BA and ED are collinear and
parallel and BC and JSFftrc noncollinear and parallel. (See the figure be
low.) The case in winch BA and ED are noncollinear and parallel and
BC and EF are collinear and parallel can be proved in a similar way.
38. Prove Theorem 7.26 for the case in which BA and ED are noncollinear
and parallel and BC and £F arc noncollinear and parallel as suggested
in the figure. (Hint. Think of the figure formed by the lines Afi, BC, DE.
and EF. Do you see that this figure contains a parallelogram? Then use
Theorem 7.15.)
39. In Exercises 37 and 38, you were asked to prove Theorem 7.26 except
for the case in which BA and ED are collinear and parallel and BC and
EFare collinear and parallel. Draw a figure for this special case. Note
that in this special case the theorem amounts to saying that equal angles
are congruent angles.
40. Draw an appropriate figure and prove Theorem 7.27 for the case in
which BA and ED are collinear and parallel and BC and EF are collinear
and antiparallel.
4L Draw an appropriate figure and prove Theorem "7.27 for the case in
which BA and ED are noncollinear and parallel and BC and EF are
collinear and antiparallel,
7.8 Parallelism for Rays 317
42. Draw an appropriate figure and prove Theorem 7,27 for the case in
which BA and ED are collinear and par&tlel and BC and EF are non
coUinear and antiparallel.
43. Draw an appropriate figure and prove Theorem 7.27 for the ease ia
which BA and ED are noncollmear and parallel and BC and £F are noo
coDinear and an tj parallel,
44. Prove Theorem 7.28. Do it by eases. In which special case does the as
sertion of the theorem amount to an assertion that vertical angles arc
congruent?
45. ABCD is a convex quadrilateral. Prove: If
m£A +m/.B + mlC + m£D = 360,
ifmZA a mZ C, andif mZB = roZ D t thenABCDis a parallelogram.
46. Prove: If Z A and Z B are consecutive angles of a parallel ©gram, then
they are supplementary. (How many pairs of consecutive angles does a
parallelogram have?)
In Kxercises 4752, write a sentence justifying the truth of tho given
statement.
47. If two alternate interior angles determ ined by a transversal of two given
copianar lines are not congruent, then the given lines are not parallel.
48. If two corresponding angles detennined by a transversal of two given
copianar lines are not congruent, then the given lines are not parallel.
49. If two consecutive interior angles determined by a transversal of two
given copianar lines are not supplementary, then the given lines are not
parallel.
50. If the diagonals of a quadrilateral do not bisect each other, then the
quadrilateral is not a parallelogram.
51. If two opposite sides of a quadrilateral are not parallel, then the quadri
lateral is not a parallelogram.
52. If ABCD Is a parallelogram, then AB and CD are everywhere
equidistant.
53. Given two alternate interior angles detennined by a transversal of two
parallel lines, prove that the angle bisectors of these angles are
antiparallel ■
54. 'Hie figure shows five copianar
lines and some (perhaps all}
of their points of intersection.
If m^L ABE + mLBEA + mLEAB
= 180, I BAE at LEAC, Z ABE
b Z DBE t and m Z BEA = 90, prove
that BD is parallel to AC.
318 Parallelism Chapter 7
7.9 SOME THEOREMS ON TRIANGLES AND QUADRILATERALS
One of the most important consequences of die Parallel Postulate
is the following theorem.
THEOREM 7«$9 The sum of the measures of tlie angles of a tri
angle is 180.
Proof* (See Figure 744,) Let A ABC be given. Let m be the unique
line through C and parallel to AB. Let D and E be points on m such that
DCE and such that B and D are on opposite sides of AC as in the
figure. Then
mLA + mLB + mlACB
= mlACD + m L BCE + mZ ACB = 1 SO.
(Which theorems justify these equations?)
It may be of interest to mention here that in the nonEuclidean ge
ometry of Lobachevsky and Bolyai the sum of the measures of the an
gles of a triangle is less than 180, The sum is not the same number for
all triangles in that geometry. The smaller the triangle is the closer die
sum is to 180.
THEOREM 7,30 The measure of an exterior angle of a triangle is
equal to the sum of the measures of its nonacljaeent interior angles.
Proof: Assigned as an exercise,
THEOREM 7,31 Let a onetoone correspondence between die
vertices of two triangles be given. If two angles of one triangle are
congruent, respectively, to the corresponding angle* of the other
triangle, then the third angles of the two triangles are also
congruent.
7.9 Theorems oti Triangles and Quadrilaterals 319
Assigned as an exercise (Note Figure 745 which suggests that
the correspondence need not be a congruence. One of the triangles
may be larger than the other.)
Figure 745
THEOREM 7.32 {The S.AJi. TJieorem) Let a onetoone cor
respondence between the vertices of two triangles be given. If two
angles and a side opposite one of them in one triangle are congru
ent, respectively, to the corresponding parts of the second triangle,
then the correspondence is a congruence.
Proof:
Assigned as an exercise.
THEOREM 7.33 The sum of die measures of the angles of a con
vex quadrilateral is 360.
Proof: Let ABCD be a convex quadrilateral as in Figure 746. Draw
diagonal BD. Since ABCD is a convex quadrilateral, D is in the interior
of Z ABC and B is in the interior of Z CDA. Then, using the notation
of the figure, we have
mLA + mZABC + mLC + mLCDA
= mZA r {mil + m/2) + m£C+(mZ4 + mZ3)
= (m IA + m Z 1 + m Z3) + m Z2 + (m LC + m Z4)
= 180 + 180 = 360,
THEOREM 7.34 The acute
complementary.
es of a right
e are
Proof: Assigned as an exercise.
320
Parallelism
Chapter 7
THEOREM 7.35 (The HypotenuseLeg Theorem) Let there be
a onetoone correspondence between the vertices of two right
triangles in which the vertices of the rigfrt angles correspond. If the
hypotenuse and a leg of one triangle are congruent to the corre
sponding parts of the other triangle, then the correspondence is a
congruence.
Proof: (See Figure 747.) Let A ABC with mZA = 90 and ADEF
with m£_D = 90 be given. Let it be given that BC^EF and
AB = DE. We shall prove that ABC * — > DEFis a congruence. You
will be asked in Exercise 19 below to supply the missing reasons,
tatement
1. Let G be a point on opp DF
such that DG s AC,
'2, mLEDG = 90
3. ED = BA and ml. A
4. ADEG =s AABC
5. EG^BC
6. BC = EF
7. EG a EF
8. ZEGDa ZKFD
9. AOEG a ADEF
10. AABCss ADEF
 <.K)
Reason
1. The Segment Construction
Theorem
2. Z £Z>G is a supplement of a
right angle
3. Given
4. S.A.S.
5. a
6. Given
7.Q]
8. E
9. S.A.A.
0. The equivalence properties
of congruence for triangles
EXERCISES 7.9
In Exercises 16, the measures of two angles of a triangle are given. Find
the measure of the third angle of that triangle,
1. 93 and 80,
% a and b. (Express the answer in terms of a and h)
3. 25 and x. (Express the answer in terms of x.)
4, 90 + k and 90 — 2k, (Express the answer in terms of Jt.)
7.9 Ttworems on Triangles and Quadrilaterals 321
5. 45 + x and 45 ■+ *. (Express the answer in terms of x.)
6. 60 + h and 60 + t. (Express the answer in terms of u and V.)
In Exercises 7ll t the measures of the three angles of a triangle are ex
pressed in terms of a number %. In each case, find the value of .v and cheek
your answer by adding the anjiie measures.
7. x, x — 5, x s 5
S. x, 2x, 3x
9. x + 1, x + 2, x + 3
10. x + 50, 2x + 30. 3x + 80
11, 3x + o»2r  3, ox — 8
12. The figure shows a triangle A ABC, a point D between A and tf, the line
CD, and the lines parallel to CD through A and B. What is the sum of
the measures of angles 1 , 2, 5, and 6? How does the sum of the measures
of Z 1 and Z 6 compare with the sum of the measures of £ 3 and Z4?
How docs the sum of the measures of / 3 and Z 4 compare with the
measure of £ACB? Use these ideas to write an alternate proof of The
orem 7.29,
13, The figure shows a parallelogram ABCD and one of its diagonals AC,
D C
What is the sum of the measures of angles LDAB and Z B? Why?
What is the sum of the measures of angles LDCB and I D?
What is the suni of the measures of the angles of a parallelogram?
Why is AB ss CV, m m Z3A~ t and 55 SM A~C?
Why is A ABC s ACDA?
Use these ideas to write an alternate proof of Theorem 7.29.
Prove Theorem 7,30,
Prove Theorem 7.31.
16. Prove Theorem 7.32. (Start with two triangles, &ABC and AA'JS'C,
and a correspondence ABC <s — > A'B'C)
It.
15.
322 Parallelism Chapter 7
17. Justify the equations at the end of the proof of Theorem 7.33.
18. Prove Theorem 7.34.
19. Copy and supply reasons for steps 5, 7, 8 in the proof of Theorem 7.35.
20. Let A ABC; and uurntan a, b. c be given. Suppose that m£A = ka,
m£B = kh, m L C = kc for some number A% the same number k in all
three cases, If a = 1,6 = 2,0 = 3, find mZA, mlB, mjLC. Check
by addition.
21. Same as Exercise 20, except that a = 2, h = 3, c ss 5.
22. Same as Exercise 20, except that a = 100, h = 250, c = 300.
23. Same as Exercise 20, except that a s 180, h = 180, a = 180.
24. Same as Exercise 20, except that fits 1, h = 1, c = 100.
25. Same as Exercise 20, except that ft = 100, b = 1 , c = 100.
■ In Exercises 2630. there is a labeled figure in which x denotes the measure
of an angle and there is an assertion about .t. Justify the assertion. In the fig
ure for Exercise 30, y and z also denote angle measures.
20, i = 50
x = 90
27. ;v = 150
29. x = 115
30. x = 27
7,9 Thwrtms wi Triangles and Quadrilaterals 323
In Exercises 3135, Ihcre is a labeled figure in which x denotes the length of
a segment. At the left, there is an assertion about x. Justify the assertion. If
ij appears in a figure, it also denotes a length.
31. x = 12
32. X = if
33. x = 10
34, X= 11
35. x = y
36. If the measure of the vertex angle of an isosceles triangle is 12.76, find
the measure of each l>ase angle.
37. If the measure of each base angle of an isosceles triangle is 83, Dnd the
measure of an exterior angle at the vertex of the triangle.
38. Find the sum of the measures of the exterior angles (one ul each vertex)
of a triangle whose angles have measures as follows: 27,32* 59.39, and
93.2&
39. Given a parallelogram A BCD with AC — BD, find the measures of the
four angles of the parallelogram.
40. Can a convex quadrilateral have three obtuse angles? Explain your
answer.
324 Parallelism Chaptef ?
41. Given a convex quadrilateral ABCD with mZA = 90, find mLB +
tnZC + mZD. Can LB and Z C both be obtuse angles?
42. Let AiJCD be aeon vex quadrilateral with m Z A = 90andmZB = 90.
Can Z C and Z D both be obtuse angles?
43. L«t AS and CD intersect at E, an interior point of each segment. If
AD = BC and AD  W, t prove that DC and AH bisect each other.
44. The measures of the angles of a triangle are 36, x t and y. We conclude
that x is a number less than [7], tliat y is a number less than {T\, and that
x + If = 13 ( In each case find the smallest number that will make the
statement true.)
45. The measures af the angles of a convex quadrilateral are a, b> c, d. We
conclude that each of these four numbers is less than [T. (Find the
smallest number that will make the statement true.)
46. The measures of the angles of a convex quadrilateral are 20, x, x, and y.
We conclude that x is a number between [T and [T. that y is a number
between £TJ and jT], (Make the smaller number as large as possible and
the larger number as small as possible in each case.)
47. Given the plane figure in which AC is the bisector of Z BAD, CB X AB S
and CD _L AD, prove that BC = DC.
48. Prove: If each angle of a convex quadrilateral is congruent to the angle
opposite to it, then the quadrilateral is a parallelogram. [Hint: Given
quadrilateral A BCD with LA^ LC % LBst Z£) } find»iZA + m/ H
and deduee that two sides are parallel. Find »«ZA + mLD and de
duce that the other two sides are parallel.)
49. Given A ABC and A A DC with JD intersecting EC at M t an interior
point of AD and EC as indicated in the figure, find the angle measures
x, y, 3 u, t\
R«view Exercises 325
CHAPTER SUMMARY
The central theme of this chapter is parallel lines, and the Iiigh point
Of I he chapter is ihe PARALLEL POSTULATE. Before introducing Lhe Par
allel Postulate we proved several theorems on conditions which imply that
lines are parallel These arc theorems in EUCLIDEAN GEOMETRY as
wdl n in the NONEUCLIDEAN GEOMETRY of Lobachcvsky and
Bcdyai. These theorems are concerned with lines and TRANSVERSALS and
angles associated with them. The Parallel Postulate gives us what we need to
prove the converses of these theorems.
PARALLELISM for lines in a given plane is an equivalence relation.
that is, it is reflexive, symmetric, and transitive. The same is true for paral
lelism of segments in a given plane and for rays in a given plane.
With parallel lines it is natural to associate PARALLELOGRAMS.
This chapter includes several theorems regarding properties of parallelo
grams and several others regarding conditions tinder which quadrilaterals
are parallelograms.
An important consequence of the Parallel Postulate is the theorem on
the sum of the measures of the angles of a triangle. An easy corollary of this
theorem is the theorem on the sum of the measures of the angles of a convex
qnadrilateraL Another important consequence of the Parallel Postulate is
the result that parallel lines are everywhere equidistant
Along with the idea of PARALLEL RAYS we introduced the idea of
A NT! PARALLEL RAYS, There were several theorems regarding the re
lationship of two angles with parallel or anti parallel sides.
Early in the book we introduced the S.A.S., A.S.A., and $.$*& Postu
lates. In this chapter there were two results which extend these congruence
ideas, namely the S.AA. THEOREM and the HYPOTENUSELEG
THEOREM for right triangles.
REVIEW EXERCISES
Copy and complete the definitious in Exercises 110.
I, line m is parallel to line n if and only if [TJ,
SL Two lines are skew lines if and only if (TJ
3, Two segments are parallel if and only if \T\.
A. Two collinear rays arc parallel if and only if jT.
5. Two noncollincar rays are parallel If and only if [fj,
& Two collinear rays arc antiparallel if and only if JT],
7. Two noncoUinear rays are antiparallel if and only if 'T].
8. If distinct lines m, n t t are coplanar, then Ms a tramversal of m and n if
and only if [TJ,
326 Parallelism Chapter 7
9. A parallelogram is a convex quadrilateral whose [?j.
10. The distance between two distinct parallel lines is [?].
■ Exercises 1 120 pertain to the geometrical figure suggested by Figure 748.
It is given that lines A& % CD, and EF arc parallel and noncoplaaar. Also,
AB = CD = EF.
In each exercise tell whether llic given statement is true or false, and explain
why it is true Off why it is false.
Figure 745 £
11. There is more than one plane contacting A and B.
12. There is more than one plane containing A, B> and C.
13. There is exactly one plane containing A, B> E, and R
14 There is exactly one plane containing A, £, C, and F.
15. Quadrilateral ABDC is a parallelogram.
16. Quadrilateral ABFE is a parallelogram.
17. Quadrilateral CDFE is a parallelogram.
18. AACE St ABDF
19. A ABC =s ADCfl
20. m / £4 C+ in lACE+m£ CEA + m L FBD +ml. BDF+ m Z DVB
= wZ CAB + m Z A£D+ mZ BDC+ m Z DCA
Exercises 2130 pertain to the geometrical figure suggested by Figure 749.
Line ( is a transversal of lines m and n, The measures of eight angles formed
by m, n, and f. have been marked in the figure. In each exercise, tell whether
tile statement is true or false. If it is true* state a theorem that justifies your
answer.
Figure 749
Rftvtaw Exercises 327
21, If * = 125, then in [\ n, 26. If m ]n, then w = 125.
22, If u = 65, then in  n. 27. If n = 60, then m is not parallel to n.
23, If w = 125, then m  n, 28. If .v = MO, then m is not parallel to n.
24, If m ' n, then 3 = 125. 29. If y =£ u, then m is not parallel to n.
25, If m  n» then u = 65. 30. x = 126
31. In your own words write a sentence that tells of a significant difference
between the geometry of Euclid and the geometry of Lobachevsky and
Bolyai.
32. Explain what we mean when we say that the equals relation is reflexive,
synunetric, and transitive.
33. Explain what we mean when we say that the relation of parallelism for
lines is reflexive, symmetric, and transitive.
34. Is the relation of parallelism for rays an equivalence relation?
;15. h the relation oi antipardllelism For rays on equiva'.ru... rln.n m.
In Exercises 3642, you are asked to prove a statement. You may use any
thing that we have had in our formal geometry structure up to this point
(except the theorem you are asked to prove) in writing your proof.
36. Prove that opposite sides of a parallelogram are congruent.
37. Prove that if two opposite sides of a convex quadrilateral are congruent
and parallel, then the quadrilateral is a parallelogram.
38. Prove that if opposite sides of a convex quadrilateral are congruent, then
the quadrilateral is u parallelogram.
39. Prove that opposite angles of a paralletogram are congruent
40. Prove that two consecutive angles of a parallelogram are supplemen
tary.
41. Prove that the two acute angles of a right triangle arc complementary
angles.
42. State and prove the theorem regarding the angle measure sum of a
triangle.
43. In the geometry of lobachevsky and Bolyai is it tnie that two alternate
interior angles formed by two parallel lines and a transversal are
congruent?
44. In the geometry of Euclid is it true that two alternate interior angles
formed by two parallel lines and a transversal are congruent?
45. In the geometry of Lohachevsky and Bolyai is it true that it' two alter
nate interior angles formed by two distinct lines and a transversal of
them are congruent, then the lines are parallel?
46. In the geometry of Euclid is it true that if two alternate interior angles
formed by two distinct coplanar lines and a transversal of them are con
gruent, then the lines are parallel?
Chapter
Fritz Henlc/Fhoto Researclien
Perpendicularity
and Parallelism
in Space
8.1 INTRODUCTION
This chapter is concerned mostly with figures that do not lie in one
plane. We use the word figure in two ways: as a set of points (geo
metrical figure) and as a picture or drawing that represents a geo
metrical figure. The figures in this book and the figures that you draw
as you study geometry are important. In communicating information
regarding geometrical figures one drawing may be worth 473 words.
When drawing a figure, the perspective view is the t>cst one from
the standpoint of communicating how the figure looks, and that is the
view used in the illustrations in this book. However, the oblique view
is the easiest to draw; therefore instructions for drawing figures in
oblique views are given on the following two pages.
A plane may be suggested by a parallelogram as in the figure
which illustrates two intersecting planes. Note that the plane sug
gested by a parallelogram includes more than a parallelogram and its
interior. It includes all of the points in the plane of lire parallelogram,
the points in the exterior of the parallelogram as well as those on the
parallelogram and in its interior.
PROCEDURES FOR
DRAWING GEOMETRIC
FORMS IN
OBlfQUE. VIEWS
Draw front plane, edge
or edges of form, in
norma* elevation view.
Piojeti from Hie
comers, al a 45*
angle, facetting
Lid'd k h'..,, wllos=
unit of measurement
sl Diid m H of those
in Iht' elevation view.
The bteck dashes
ore guide lines to
b* crated later.
z
Draw vertical gu.de
lm« for the center
of tne pyramid.
/ /
y
y* 76
330
A
7 /
£JZ7
/
ines that aro Inside
or bor.irtd an obfect
are by convention
drawn 35 dashes.
331
332 Parpandlcularity and Parallelism Jn Space
Chapter 8
A horizontal plane is oriented like the top of the cube. A vertical
plane is oriented like the right face of the cube. Note Lhat horizontal
and vertical are not defined formally in our geometry. They arc
descriptive terms that we use to describe figures.
The main topics in this chapter are perpendicularity and parallel
ism for lines and planes, in the preceding chapters of this book we
have given detailed proofs for most of the theorems that were stated.
Proofs not included in the text were assigned as exercises. In this
chapter, however, we sometimes state theorems without giving proofs
and without assigning these proofs in the exercises. In such cases we
may include an outline of the main steps or a short statement of the
strategy or main idea of a possible proof. In all cases it is possible to
write detailed proofs based on the postulates of our formal geometry.
Our abbreviated presentation is appropriate for a first course in formal
geometry.
8.2 A PERPENDICULARITY DEFINITION
Definition 8.1 A line and a plane are perpendicular if the
line intersects the plane and is perpendicular to ever)' line in
the plane through the point of intersection.
If a line / and a plane a are perpendicular, we say that I is perpen
dicular to a and that a is perpendicular to /.
Notation, / 1 a, or a 1 /, means that a and / arc perpendicular;
I J_ a at P, or a 1 2 at P, means that t X a and that P is their point of
intersection.
Figure 81 represents a line / and a plane a that are perpendicular
at P. The figure shows four of the lines through P and in a. Actually,
there are infinitely many such lines and they are all perpendicular to
I at P.
riffureSi
9.2 A Perpendicularity Definition 333
EXERCISES 3.2
■ In Exercises 19, draw the figure that is described by the following
statements.
1, Two distinct horizontal planes,
2. Two distinct vertical planes.
& A horizontal plane and a vertical line,
4, A vertical plane and a horizontal tine.
5. A vertical plane and a line perpendicular to it
6* Two distinct parallel planes and a line that intersects both of them,
7. Two distinct parallel lines and a plane that is parallel to both of the lines.
8. Two distinct intersecting planes, one vertical plane and one horizontal
plane.
9. Two congruent triangles tying in distinct parallel planes and the seg
ments connecting corresponding vertices,
10, See the figure. If i _L m, and m lies in a. docs it follow that / J_ a?
Rxplain
1L In the figure, ABA\ AB = A% aX' I a at B t B $6 C, and C £ a.
Prove that
AC = A'C
A
\
! /
A'
334 Perpendicularity and Parallelism in $p*ce Chapter 8
12, In the following figure, aA' j. a at P, A'PA, A? = A'P, and BDC.
On the basis of Exercise 1 i f what can you say about the lengths AB and
A% AC and A'C, AD and A'D? Now prove that AABD = AA'BD,
AACD s= AA'CD, A ABC s AA'JSC.
13. In the figure in Exercise 12, AT ± PB, AP ± PC; A'PA; BDC; a
is the plane containing the noncollinear points B, C, P; AB — A'B;
AC = A'C. State the postulate which implies that D is in a, Prove that
AD := A'D.
14. {An informal geometry exercise*) Let I and m be two distinct lines in
the plane of a table top and intersecting at a point R Hold a yardstick
so that it appears to be perpendicular to I and to in at P. Use another
yardstick to represent a third line n in the plane of the table top and
passing through P. Docs the first yardstick appear to be perpendicular
to lino n? Try n in various positions and see if you can find a position
such that the yardstick and line n are no longer perpendicular. Does it
appear that the yardstick is perpendicular to lino n in all cases? Does it
appear that the yardstick is perpendicular to the table?
8.3 A BASIC PERPENDICULARITY THEOREM
Theorem 8.1 is introduced to help us prove a basic perpendicularity
theorem, Theorem 8.2. Theorem 8.2 is suggested by our experiences
with perpendicular lines and planes. See Exercise 14 of Exercises 8.2,
THEOREM 8. / If A, A ', B, C, D are distinct points with B and C
each equidistant from A and A' and with D on BC, then D is equidis
tant from A and A f .
Given distinct points A, A\ B, Q A with AB = A'B, AC =
A'C, and D on BC as suggested in Figure 8.2, we want to prove that
AD s A'D. Then DBQ or BDC 7 or BCD. The proof may be
completed by showing that A ABC ^ AA'BC* AABD =* AA'BD*
AACD a AA'CD. and hence that AD = A'D*
8,3 A Basic Perpendicularity Theorem 335
qp»u
THEOREM 8.2 If a line is perpendicular to each of two distinct
intersecting lines at their point of intersection, then it is perpen
dicular to the plane that contains them.
Proof: Let lines l r m T n be given with m, n in plane a and such that
I _L m at A y I 1 natA, as suggested in Figure 83. Let p be any line
distinct from TO and n that ties in a and passes through A. We want to
prove that I 1 p. Let q he a line in a that i s not parallel to m, or to n, or
to p. and that docs not pass through A. Then q intersects to, p* rt in
three distinct points; call them M, P, N, respectively.
Let Q awl (r be two points of I such that QAQ' and QA = Q'A,
The proof may be completed by showing that M is equidistant from
Q and Q\ S is equidistant from Q and Q', P is equidistant from {)
and Q'j
AQPA a AQ^PA, LQAP^ LQ'AP, and I X p.
Since p is an arbitrary line oilier than to and n in the plane a and pass
ing through the point A, we have I X a.
Vigun W
336 Perpendicularity and Parallel ism in Space
Chapter 8
EXERCISES 8.3
1. Let the distinct points A, B, C, A E be given as suggested in the figure.
A, B r C. D are coplanar points with no three of them eollinear. Also
ED I W and ED 1 DC, Prove that ED 1 M.
.?*
2. Let the distinct points A, H> C t D be given as suggested in the figure
A, /*, C are noncollinear points in a plane «, S3 I a, and JJ = #C,
Prove that IDAC a ZDCA,
3. Let the distinct points A, S t C, £> be given as suggested in the figure,
ATS 1 J57), and AB _L HC. A, B, C are noncollinear points is a, and I)
4—*
is a point not in a. Prove that AD is not perpendicular to a.
8.3 A Basic Perpendicularity Theorem 337
4, For the situation in Exercise 3, prove that AD is not perpendicular to
tlie plane determined hy B, C D. Is AjB perpendicular to the plane
BCD?
5, Let four noncoplan&r points A> B, C, D be given such that
AB = AC = AD,
A3 i plane ACD, AC 1 plane BAD, A& X plane BAG. Draw an ap
propriate figure and prove that A BCD is an equilateral triangle.
6, A rectangular solid has the property that if two of its edges intersect,
then they intersect at right angles. The following figure shows a rcctan
gular solid with the eight vertices labeled. Prove that EA is perpendicu
lar to the plane that contains B t C, D.
7, Assume the same situation as in Exercise 6. Since EA is perpendicular
to the plane that contains the quadrilateral ABCD t wc say that edge EA
is perpendicular to face A BCD. Name six other combinations of an edge
and a face that are perpendicular. (There are 24 such combinations
altogether.)
8, In the proof of Theorem 8.1, suppose that DBC. Which Congruence
Postulate would you use to show that AABCs AA'BC? That
AABD Sf AA'BD?
9, Explain how the proof of Theorem 8.1 for the case in which BDC
differs from the proof for the case in which DBC, Draw an appropri
ate figure.
10. Draw a figure like Figure 83, except much larger. Draw the segments
connecting Q and Q with M, P, N. In completing the proof of Theorem
8.2, which theorem, postulate, or definition plays a key role in proving
that
(ft) M is equidistant from Q and Q' and that N is equidistant from Q
and Q'?
(b) F is equidistant from Q and p?
(c) &QPA s AQ'PA?
(d) LQAF=z Zp'AP?
(e) / 1 p?
338 Perpendicularity and ParaKtali&m in Space
ChaplerS
8,4 OTHER PERPENDICULARITY THEOREMS
THEORFM 8,3 If a line and a plane are perpendicular, then the
plane contains every line perpendicular to the given line at the
point of intersection of the given line and the given plane.
Proof: {See Figure 84.) Let / I a at P. Let t x be any line such that
/, ^/atP.
Figured
We want to prove that ^ lies in «, Let be the plane that
contains / and ly (See Figure 83.) Then /? ^ a since contains / and
g does not. The intersection of (i and a is a line; call it / 2 . Then 1% _L J.
Why? Also h ± I Why? Then h = k* Why? Therefore /, lies in a.
Since ^ is any arbitrary line perpendicular to / at F, die proof is
complete.
figure. S5
8,4 Other Perpendicularity Theorems 339
THEOREM 8.4 Given a line and a point, there is a unique plane
perpendicular to the line and containing the point
Proof: Lei £ be a given line and P a given point. We divide the proof
into four parts.
1. If P £ /, we must prove that there is
at least one plane perpendicular to /
at P.
2. If P £ J, we must prove that there is
at most one plane perpendicular to I
at R
3. If P £ /, we must prove that there is
at least one plane perpendicular to I
and passing through P.
4. If P £ I, wc must prove that there is
at most one plane perpendicular to I
and passing through P.
Parts 1 and 3 are existence proofs; parts 2 and 4 are uniqueness
Proof of 1; If P is a point on a line 2 as in Figure 86, let rt be a plane
containing I and let in be the line in a that is perpendicular to I at P.
Figure Wl
I^t Q be a point not in a and let fi be the plane that contains Q and L
Let n be the line in /J that is perpendicular to / at P. Let v ta the plane
that contains m and n. Then I A. y at E Why? This proves that there
is at least one plane perpendicular to 1 at P.
340 Ptrpandteularity and Parallelism In Spac*
Chapters
Proof of 2: If point P lies on a line I as in Figure 87, we shall prove
that there is at most one plane perpendicular to / at P. Suppose, eon
Figure S7
trary to what we assert, that there are two distinct planes a and /? each
perpendicular to I at P. Then a and /? intersect. Why? Their intersection
is a line. Why? Call this line m. Let f and s be the lines such that r is in
a and rimat F, and * is In /? and s 1 m at P, Let y be the plane that
contains r and s. Then m J_ y at P (Why?), m X I at f (Why?), and I
lies in y. Then % r, s are three distinct eoplanar lines with r _ I and
s J. I, But this is impossible, There cannot be two distinct planes each
perpendicular to I at P. Therefore there is at most one plane perpen
dicular to I at P,
Proof of 3: Given a line / and a point P not on I as in Figure 88, let
Q be the foot of the perpendicular from P to I Let a be the unique
plane such that a ^_ I at Q. How do we know that there is one and only
one such plane a? Then P lies in a. Why? Therefore there is at least
one plane containing P and perpendicular to I .
Figure B9
8.4 Other Perpendicularity Theorems 341
Proof of 4: Given a line I and a point P not on it, as in Figures 89
and 810, we shall prove that there is at most one plane perpendicular
to / and containing P. Suppose, contrary to our assertion, that there
are two distinct planes a and ft each perpendicular to / and each con
taining P. Then the intersection of a and {S is a line; call it m.
Figure S9
Suppose I intersects m as in Figure 89. Then I intersects m in a
point Q different from P. Why? Then a and /3 are each perpendicular
to I at Q t But this is impossible. Why? Therefore I does not intersect m.
Therefore there are two distinct points Q and K, as in Figure 810,
such that I ± a at Q and I 1 £ at R. Then ARPQ has two right angles.
Figure HIQ
But this is impossible. Why? Therefore there cannot be two distinct
planes each perpendicular to I and each containing P. Hence there is at
most one plane perpendicular to I and containing P.
Definition 8.2 If A and B are distinct points, the unique
plane that is perpendicular to AB at the midpoint of SB is
called the perpendicular bisecting plane of AB.
342 Perpendicularity and Paralleflwt In Space
Chapter 8
THEOREM 8.5 The perpendicular bisecting plane of a segment
is the set of all points equidistant from the endpoinls of the segment.
Let a be the perpendic
ular bisecting plane of A B and let
C be the midpoint of A5. (See
Figure 811.) Let P be a point.
There are two things Lo prove,
1. IfPCa J thcnAP = ¥R
%, If AP = PB,thenf€a,
You are asked to prove these two
statements as exercises.
Figun* SI 1
THEOREM 8,6 Given two perpendicular lines, there is a unique
line that is perpendicular to each of the given lines at their point of
intersection.
Proof: Let I and m be lines that are perpendicular to each other at P
as in Figure 812. l^et a be the unique plane that is perpendicular to
I at P, Let n be the unique line in a that is perpendicular to m at P.
Then n is also perpendicular to L Why? Therefore there is at least one
line perpendicular to both I and m at their point of intersection.
Figure SI2
Suppose that it and n' are distinct lines each perpendicular to m
and to I at P as in Figure 813.
Figure S13
8,4 Other Perpendicularity Theorems 343
Let a be the unique plane that is perpendicular to I at P. Then n, n',
m are three distinct lines in a and n 1 m, n' _L m. Since this is impos
sible, there is at most one line perpendicular to both I and m it then
point of intersection.
Since there is at least one line and at most one line that is perpen
dicular to / and in at P t it follows that there is one and only one line
that is perpendicular to both I and m at their point of intersection.
THEOREM 8.7 If two lines are perpendicular to the same plane,
they are parallel.
Let I and m be lines that are perpendicular to plane n\ as in
Figure 814 If I = m, then l\,m and there is nothing more to prove.
Suppose, then, that I ^ m. Then it can be shown that I and m are non
intersecting lines. (If / and m intersect at a point A not in a, then A
and the two points where I and m cut ct are the vertices of a triangle
with two right angles. If I and m intersect at a point A in or, then the
plane containing I and m intersects a in a line n such diat n ll,ni m,}
Figure S14
1 *t F and Q be the points in which / and m, respectively, intersect
a. Suppose, contrary to what we shall prove, that / and m are skew lines.
I^t m* l>e the unique line through Q and parallel to I. Then / _L PQ
and m* _ PQ. Why? Let t and & be die unique lines in a that are per
pendicular to r\) at F and at Q f respectively. Then I ± r and it fol
lows from Theorem 7.26 that m' _L ,v, Since m' J_ PQ and m* I 5, it
follows that m' _L a. Let }i be the plane that contains m and »i' and
let t be the line in which /? and a intersect. Then m _L £ m' J_ (, and
the three distinct lines m f m', t are eoplanar. xSince this is impossible,
it follows that I and m are not skew lines* Therefore they are coplanar
nonintcrsecting lines, that is, they are parallel.
344 Perpendicularity and Parallelism in Space
THEOREM S.8
Chapter 8
If one of two distinct parallel lines is perpendicu
lar to a plane, then the other line is also perpendicular to that plane.
Proof: Given two distinct parallel lines / and m and a plane « per
pendicular to l t let P and Q be the points in which / and m intersect tx t
as suggested in Figure 815, Let r s PQ, let s be the unique line in a
perpendicular to r at Q, and let mf be the unique line perpendicular
to r and s at Q (see Theorem 8,6). It follows from Theorem 8.2 that
m' is perpendicular to or., from Theorem 8.7 that m* is parallel to h,
and from the Parallel Postulate that m' = m. Since m f is perpendicular
to a, it follows that m is perpendicular to a.
•I
Figure S15
THEOREM 5,9 Given a plane and a point, there is a unique line
containing the given point and perpendicular to the given plane.
Proof: Let a plane a and a point P be given. We shall prove that there
is a unique line through P and perpendicular to a. We divide the proof
into four parts: two existence proofs 1 and 3 and two uniqueness proofs
2 and 4.
1. If P € a, we must prove that there is
at least one line perpendicular to or
at P.
2. If P £ <*. w c must prove that there is
at most one line perpendicular to at
at P.
3, If P g a, we must prove that there is
at least one line perpendicular to ot
and containing P.
4, If P g a, we must prove that there is
at most one line perpendicular to a
and containing P.
8,4 Other Perpendlculirity TTworems 346
Proof of 1: Given a plane a and a point P in a as suggested in Figure
816, let m and n be two lines in a that are perpendicular to each other
at P. Then it follows from Theorem 8.6 that there is a unique line I that
is perpendicular to both m and n at P. Then I _ a at P. This proves that
there is at least one line perpendicular to a at P.
Figures 1(1
Proof of 2; Given a plane a and a point P in a, as suggested in Figure
817, suppose contrary to what we shall prove, that m and n are dis
tinct lines and that each is perpendicular to a at P. Let /? be the plane
that contains m and n and let I be the line in which /? intersects «. Then
m l/,nl f, and the three lines I, m, n arc coplanar. This is impossible.
Therefore there cannot he two distinct lines through P each perpen
dicular to a.
Figure SI
Proof of 3; Given a plane a and a point P not in a, as suggested in
Figure 818, we shall prove that there is at least one line through P and
perpendicular to a. Let Q be any point of a. If PQ J a, there is noth
ing more to prove. Suppose, then, that FQ is not perpendicular to a.
346 Ptrjwndicularity and Parallelism In Sf»c«
Chapter 8
Let I be the unique line such that IX a at Q. How do you know there
is one and only one such line J? Let m he the unique line through P and
parallel to /, Then it follows from Theorem 8.8 that m L <*. Therefore
there is at least one line through P and perpendicular to a.
Figure 848
Proof of 4: Let a plane a and a point P not in a he $ven. If there
were two distinct lines through F each perpendicular to a, then there
would be a triangle with two right angles. Since this is impossible, it
follows that there is at most one line through P and perpendicular to a.
EXERCISES S.4
1. Let four noncoplanar points A, B, C, D such that KB at "KC S6 AD,
KB _L AC) AB _L A75, A7? JL ]S75 he given. Draw an appropriate figure
and prove dial A BCD is not a right triangle.
2. I ,et four noncoplanar points A , B, C, D such tliat AB _L AT7, KB JL AD,
7VC _L AD be given. If A? is the bisector of L BAC, prove that / DAE
is a right angle.
3. Let t m, n be three distinct lines, not necessarily coplanar, and such that
1 1 m and m j n. Use Theorems 8.4, 8.7. and 8.8 to prove that I j n.
(This proves that parallelism for lines is a transitive relation.)
4. Let A, B, C t D, E, F be distinct points such that ABC, AB = BC,
AD = DC, AE = EC, AF = FC. Draw an appropriate figure and ex
plain why B f E, D, F are copianar points.
5. Let segments AS and UD and three points E, W t G be given. If AT? and
UD are perpendicular and intersect at E, if FEG and FT. L AB, docs
it follow necessarily that FG is perpendicular to plane ABC? Draw an
appropriate figure.
6. \jet segments AB and C/5 and three points £. F, C be given. If AB and
CD are perpendicular and intersect at E, if FEG, TG* X AT?, and
FD _ CD, does it follow necessarily that FG is perpendicular to plane
A BC ? Draw an appropriate fiu,i IE&
8.5 Parallelism for Lines and Planes 347
7. In part 2 of the proof of Theorem 8.4, how do wc know that the inter
section of a and ft is a line?
8. In purl 2 of the proof of Theorem 8.4, how do we know that ml y at P?
9. In part 3 of the proof of Theorem 8.4, how do we know that P lies in a?
10. Given two distinct points A and B in a plane a describe the set of all
points in a that arc eqtridistant from A and B.
11. Givon two distinct points A and B describe the set of all points that are
equidistant from A and B.
12. Given a line / and the set § of all lines parallel to /, prove that if F is any
point in space, then P lies on one and only one of the lines in $.
13. Give a plane a and the set § of all lines perpendicular to a, prove that
if P is any point in space, then P lies on one and only one of the lines
ing,
14. See the proof of Theorem 8.5. Prove that if P £ a y then AP = PB.
15. Sec the proof of Theorem 8,5. Prove that if AP = PB, then P £ a.
IG. Given A ABC and two points DandE such that BA 1 AB, UA 1 AC,
EE 1 AB : EB 1 EC, prove that A E, A, B are coplanar points,
17. See the proof of Theorem 8,9. In part 4 of this proof wc asserted that if
there were two distinct lines through P each perpendicular to «, then
there would be a triangle with two right angles. Draw an appropriate
figure and prove this assertion.
8.5 PARALLELISM FOR LINES AND PLANES
In this section we shall investigate the properties of planes parallel
to planes and of lines parallel to planes. Wc begin with two definitions.
Definition 8.3 Two planes are parallel if their intersection
is not a line.
Definition 8,4 A line and a plane are parallel if their inter
section is not a point.
Let us consider Definition 8.3 first, If a and ft arc planes, then the
intersection of a and ft is (1) the null set, or (2) a line, or (3) a plane.
If* n ft = t then a^=ft and a \\ft.
If a n ft is a plane, then « H ft = a = ft and a  ft.
If a n ft is a line, then a is not parallel to ft.
Therefore a is parallel to ft if and only if
a = ft or a = .
348 Perpendicularity and Parallel*™ in Sface
Chapter 8
It follows immediately from Definition 8,3 that parallelism for planes
is reflexive and symmetric. Later we shall see that parallelism for planes
is also transitive, and therefore an equivalence relation.
The fact that parallelism tor lines is an equivalence relation follows
from Definition 7.1, and Theorems 8.4, 8.7, and 8.8. See Exercises 10,
11, 12 of Exercises 7.6 and Exercise 3 of Exercises 8.4.
Now let us consider Definition 8.4. Tf Ms a line and a a plane, then
the intersection of I and of is (1) the null set, or (2) a point, or {3) a line.
If I fl a = , then / 1 1 «. If / fi a is a line, then that line is t and 1 1 a.
Ul n a is a point, then I and a are not parallel. Therefore I , a if and
only if I C a or t H a = 0,
THEOREM 8.10 If a plane intersects one of two distinct paral
lel hues but does not contain it, then it intersects the other line and
does not contain it.
Proof: (See Figure 819,) Let I and m be two distinct parallel lines.
Let a be the plane that contains / and m, I^et /? be a plane that inter
sects one of the lines, say I, in a single point P. Now P lies in both ft
and a. Since a contains / and $ does not contain /, it follows thai a =£ ft.
Therefore the intersection of ft and a is a line; call it n. Since n lies in ft
and I does not lie in ft t it follows that n is different from I. Also, m does
not lie in ft. (Suppose m lies in ft. Then m lies in a and ft, and m = n.
Since n is not parallel to l t it follows that m is not parallel to L Con
tradiction. Therefore m does not lie in ft,)
Figure S19
Therefore m, n, I are distinct lines in plane a with I parallel to m
and with n intersecting I in a single point. It follows from the Parallel
Postulate that n intersects m in a single point; call it Q. (Suppose n
does not intersect m. Then n and £ are two distinct lines through P and
parallel to m. Contradiction.) It follows that ft contains Q and hence
that ft intersects m in a single point
8.5 Parallelism for Lints and Planes 349
THEOREM S. II If a plane is parallel to one of two parallel lines,
it is parallel to the other also
Proof: Assigned as an exercise,
THEOREM 8.12 If a plane intersects two distinct parallel planes,
the intersections are two distinct parallel lines.
Proof: Let a and ft be two distinct parallel planes and y a plane that
intersects both of them, as suggested in Figure 820. Then y is distinct
from a and from fi t and it intersects each of them in a line. Let / and m
be the lines in which y intersects a and ft t respectively; then I and m
arc distinct coplanar lines which do not intersect. Why do they not
intersect? Therefore I and m are distinct parallel lines.
Figure 820
THEOREM 8.13 If a, ft y are th ree distinct planes si ich that /? is
parallel to y and such that a intersects /?, then a intersects y.
Proof: (See Figure 82 L) Suppose that a, /?, y are distinct planes, that
/? and y are parallel, and that a intersects 0,
Figure 821
350 Perpendicularity and Parallelism In Space
ChaptBf 8
Suppose, contrary to the assertion of the theorem, that a does not
intersect y. Let m he the line in which a and /? intersect. Let P be a
point of m; let I be a line through P that lies in a but not in ft and R
a point of y. Let 8 be the plane containing I and R, Then 5 intersects $
in a line n distinct from m, and B intersects y in a line r. Then / and n
are distinct coplanar lines through P and parallel to r. Since this con
tradicts the Parallel Postulate, it follows that a intersects y.
THEOREM 8,14 If a line intersects one of two distinct parallel
planes in a single point, then it intersects the other plane in a single
point.
Proof: Let a and ft be two distinct parallel planes and let I be a line
that intersects a in a single point; call it F, (See Figure 822.) Let y be
any plane that contains I, Then y and a arc distinct intersecting planes.
Their intersection is a line; call it m. It follows from Theorem 8.13 and
Theorem 8.12 that y intersects /? in a line; call it n. Then m and n arc
distinct parallel lines that arc coplanar with I. Since / intersects m and
is distinct from it, it follows from the Parallel Postulate that {intersects
n and is distinct from it. Therefore the intersection of I and n is a single
point; call it Q Since Q is a point of n, it is also a point of f$. Therefore
Q is the unique point in which / and j8 intersect
Figure 822
THEOREM H. 15 If a line is parallel to one of two distinct parallel
planes, it is parallel to the other plane.
Proof: Assigned as an exercise.
THEOREM 8 JO There is a unique plane that contains a given
point and is parallel to a given plane.
8.5 Parallelism for Line* and Plane* 351
Proof: (We give only a plan for a proof.) Let a point P and a plane or
be given. Divide the proof into four major parts.
1. If P is in a, explain why there is at least one plane containing P
and parallel to a.
2. If P is in a, explain why there is only one plane containing P and
parallel to or.
3. If P is not in a, let / he the unique line through P and perpendicu
lar to a . Let ft be the unique plane that is perpendicular to I at P,
Prove that a j /?,
4. If P is not in a, prove that there is at most one plane through P
and parallel to a.
THEOREM 8.17 Given a point and a plane, then every line con
taining the given point and parallel to the given plane lies in the
plane containing the given point and parallel to the given plane.
Proof: (Plan only.) Let a point P and a plane a be given. Suppose first
that P lies ma. Then a itself is the unique plane containing P and par
allel to «. Show that the assertion of the theorem is true in tins case.
Suppose nest that P is not in a. Let /? be the unique plane contain
ing P and parallel to «. Then a=£fi. I.et I be any line through P and
parallel to a, as suggested in Figure 823. Use Theorem 8.14 to com
plete the proof.
Fi'gure S2rs
EXERCISES SJ
L If a and j$ are planes such that or n fi = 0, is n f /J? Why?
2. If / is a line and a is a plane and !n«= 0, is it possible that (a) t is
parallel to a? (b) / is not parallel to a? Explain.
3* Let a; /?, y be three distinct planes with a parallel to # and a not par
allel to y From which theorem of this section may we conclude that $
is not parallel to y?
352 Perpendicularity and Parallelism In 5p*c* Chapter 8
4. Prove Theorem SJLt.
5. Given a plane a and two distinct parallel lines I and m such that I is not
parallel to a. From which theorem of this section may wc conclude that
m is not parallel to a?
6. See the proof of Theorem 8.13. Draw a figure like Figure 821. Add some
segments and a label to the figure to suggest the plane &
7. Prove Theorem 8.15.
8.6 PARALLELISM AND PERPENDICULARITY
hi tins section we define measure of a dihedral angle, right dihedral
angle, and perpendicular planes. This section includes, as you might
have guessed, several theorems on parallelism and perpendicularity for
lines and planes. Wc begin by repeating a definition from Chapter 4,
Definition 4.20 If two noncoplanar halfpknes have the
same edge, then the union of these halfpknes and the line
which is their common edge is a dihedral angle. The union of
this common edge and either one of these two halfpknes is a
face of the dihedral angle. The common edge is the edge of
the dihedral angle.
Figure 824 suggests two dihedral angles, one of them appearing to
be larger than the other. What does "larger" as used here mean? In
the case of (plane) angles wc defined larger in terms of measure. Since
we do not have a measure for a dihedral angle, we develop one.
Figure S24
Definition 8,5 The intersection of a dihedral angle and a
plane perpendicular to its edge is a plane angle of the
dihedral angle.
8.6 Parallelism and Perpendicularity 353
Figure $25 suggests a dihedral angle ABCD, a plane a such that
BC _i_ a, and L PQR, a plane angle of the dihedral angle ABCD.
Figure sas
It seems plausible that all plane angles of a given dihedral angle
should he congruent. This brings us to the next theorem.
THEOREM 8. 18 Any two plane angles of a dihedral angle are
congruent.
Proof: Suppose that I ABC and LDEF
are two plane angles of dihedral angle
(2—HlJ t as in Figure 826', We may suppose
mat points have been picked in these plane
angles and then labeled so that A and D lie
in the plane GHI, C and F lie in plane IHJ,
and
AB = BC = DE = EF.
It is easy to show that BCFE and BADE are
parallelograms, that ACFD is a parallel
ogram, that AABCs A DBF, and hence
IABC& LDEF,
With Theorem 8.18 proved, it is now possible to define a measure
for dihedral angles.
Definition S. 6 The measure of a dihedral angle is ihe meas
ure of any one of its plane angles.
Perpendicularity and Parallelism in Spact
Chapter 8
Definition 8. 7 A right dihedral angle is a dihedral angle
whose measure is 9G>
Definition 8.8 Two planes are perpendicular if their union
is the union of four right dihedral angles.
THEOREM 8. 19 If a line is perpendicular to a plane, then any
plane containing the given line is perpendicular to the given plane.
Proof: Assigned as an exercise,
THEOREM 8.20 If a line is perpendicular to one of two parallel
planes, then it is perpendicular to the other plane also,
Proof: Assigned as an exercise.
THEOREM 8.21 If two planes are perpendicular, then any line
in one of the planes and perpendicular to their line of intersection
is perpendicular to the other plane.
Proof: Assigned as an exercise.
TfTEOREM 8,22 If two distinct intersecting planes are perpen
dicular to a third plane, then their line of intersection is perpen
dicular to the third plane.
Proof: Figure 827 suggests two distinct intersecting planes a and fi
perpendicular to plane y. Let I = AB be the line of intersection of a
and fii let EC and BD he lines in which m and ft respectively, inter
sect v. Let h be the line in a that is perpendicular to BC at B.
Figure 827
8.6 Parallelism and Perpendicularity
Let ?2 be the line in /3 that is perpendicular to SD at B, It follows from
Theorem 8,21 that h ± y and l 2 i y. The proof may be completed
by showing that Ii ss% a 1, and hence that / J. y<
Definition &9 A segment, or ray, is perpendicular to a
plane if the line which contains it is perpendicular to the
plane, If a segment is perpendicular to a plane and otie end
point lies in the plane, then that segment is a perpendicular
to the plane, and its endpoint in the plane is the foot of the
perpendicular.
Definition 8.10 If « is a plane and S is a set of points, then
the projection of S on a is the set of all points (>, each of which
is the foot of the perpendicular from some point of S,
THEORFM 8.23 The projection of a line on a plane is either a
line or a point.
Iet t lie a line and a a plane. There are three cases to consider.
1. /lies in cr.
2. / is perpendicular to a.
3 / is not perpendicular to a and does not lie in <*.
PrcH>fofCase 1: If J lies in a t then every point of / is its own projection
on a t and therefore J is its own projection on a.
Proof of Case 2: If / is perpendicular to a at the point Q, then the pro
jection of every point of t on a is the point Q, and therefore die projec
tion of / on a is the point Q\
Proof of Cam 3: If / is not perpendicular to or and docs not lie in a T
let A, B, C be three distinct points of / that are not in a, and let A', B',
C be their respective projections on a. It follows from Definition 8. 10
that AA' ± «, BB' J_ a, CC  a, and from Theorem 8J that AA' 
BB\ BB' \ CC, AA' CC< Since / is not perpendicular to a and since
A, B, C are distinct points, it follows that AA' r BB', CC* are distinct
lines, T^t /3 be the plane that contains AA' and BB'. Then /3 contains
the point C. Why? Since /? is parallel to S3' (Why?) and M' Is parallel
to CC7, it follows from Theorem 8.11 that jS is parallel to £<?. Since fi
contains C and is parallel to CC7, it follows that jS contains CC 1 . There
356 Perpendicularity and Parallelism In Space Chapter S
fore C lies in fi. Let f be the line in which /? and a intersect. Since
A', B\ C all lie in a and in fi, it follows that they all lie on V. If / in
tersects a t say in point D, then D is its own projection on a, and since
D lies in both a and /•?, it follows that D lies on V ,
Note that I is determined by A and B, and that V is determined by
A' and B*. We have shown that if P is any point of I distinct from A and
B (like C and D in the preceding paragraph), then its projection F on a
lies on t.
Conversely, if Q' is any point of l\ then Q* is the projection of some
point Q on (, If I and a intersect in {X then Q? is its own projection
and Q s £7. If Q' is a point of P not on L then there is a unique per
pendicular to a at Q' ? and this perpendicular intersects ( in the point
Q which has the point Q* as its projection on or. It follows that every
point of / has some point of V as its projection, and that every point of
r is the projection of some point on I. Therefore the projection of I on
« is the line V,
COROLLARY 8.23.1 The projection of a segment on a plane is
either a point or a segment.
Proof: Let s be a segment, or a plane, and $' the projection of s on a.
If s is perpendicular to «, then *' is a point If s lies in a r then s' = s
and hence s' is a segment. Let I be any line not in a and not perpen
dicular to a. Let A, B, Che three points of I such that ABC and let
A\ R, C be their respective projections on a. Then it follows from
Theorem 8.23 that A' } B\ C are collinear and from Theorem 8,7 that
A'R'C. (IfA'B'C were not true, then we would have distinct par
allel lines that intersect,) Since ABC on I implies A'B'C, we say
that betweermess for points is preserved in the projection from ion a,
If s = AC and s' = WU, then B is between A and C if and only if ff
is between A' and C. Therefore s' is the projection of s.
THEOREM 8.24 Given a plane a and two distinct points A and
B su ch tha t AB is parallel to a t if AW is the projection of XB on a,
then A'W^AB.
Frtxtf: Assigned as an exercise,
THEOREM 8.25 Given parallel planes a and /? and A A RC in a,
if A', B', C are the projections of A, B, C, respectively, on j5,
then AARC^&A'B'C.
Proof: Assigned as an exercise.
8.6 Parallelism and Perpendicularity
THEOREM 5.26 The shortest segment joining a given point not
in a given plane to a point in the given plane is the perpendicular
that joins the given point to its projection in the given plane.
Roof: Assigned as an exercise.
THEOREM 8.27 All segments that are perpendicular to each of
two distinct parallel planes and have their endpoints in these planes
have the same length.
Proof: Assigned as an exercise.
Definition 8.11 The distance between a point and a plane
not containing it is the length of the perpendicular segment
joining the given point to the given plane.
Definition 8.12 The distance between two distinct parallel
planes is the length of a segment that joins a point of one of
the planes to a point of the other plane and is perpendicular
to both of them.
EXERCISES 8.6
1. There is no definition of berweenness for half planes in this book. Write
your own definition for it. Using your definition, prove that if 3C lt 3C 2 ,
X3 arc half planes with 3C Z between 0Ci and 3Ca, then the measure of
the dihedral angle formed by 5C X and 3C 3 and their common edge is the
sum of the measures of the dihedral angles formed by 3C j and 3C2 and
their edge and by 3Ca and 3Cs and their edge.
2. There are no definitions in this book for Ihc interior and the exterior of
a dihedral angle. Write your own definitions for these terms.
3. Draw a picture of a cube and label its vertices. How many right di
hedral angles are there each containing two faces of the cube?
4. Given the labeled cube of Exercise 3, select one of the right dihedral
angles and, using the vertices of the cube, identify two of its plane
angles.
5. See the proof of Theorem 8.18 Draw a figure like Figure 8*26 and mod
ify it to show the quadrilateral ACFD. Prove that ACFD is a rectangle.
358 Perpendicularity and Parallelism In Spaca Chapter 8
B Exercises 616 are concerned with projections on a plane. In Exercises 614,
sketch one or more figures and he prepared to defend your answer.
0, Is the projection of every triangle a triangle?
7. Is the projection of every ray a ray?
H. Is the projection of every point a point?
9. Is the projection of every segment a segment?
10, Ls the projection of every line a line?
1 1. Is there an angle whose projection is a line? \ ray? A segment? A point?
An angle?
12. ls there an acute angle whose projection is an obtuse angle?
13, Is there a right angle whose projection is an acute angle? An obtuse an
gle? A right angle?
14. Is there a segment that is shorter than its projection? linger?
15, If an edge of a cube is perpendicular to a plane, describe the projection
of the cube on the plane. 1 5 raw a sketch,
10, challenge problem. If a diagonal of a cuIhj is perpendicular to a
plane, draw a sketch of the projection of the cube on the plane.
■ In Exercises 1723, draw an appropriate figure and prove the theorem,
17. Theorem 8,19 21. Theorem 8.25
18. Theorem 8.2G 22. Theorem 826
19. Theorem 8.21 S3. Theorem 8.27
20. Theorem S24
24. In the proof of Theorem 8,22, show that h = fe = '■
25. Let ABCD be a right dihedral angle with M 1 B€, EC 1 CD,
AB = 3 v#, BC = 5, CD = 12, Prove that AD = 14.
26. The figure shows four noncoplanar points A, B, Q D such that
AB _L AC, AB J. AT5. XC I AD, AB = AC = AD.
Find the sum of the measures of I BDA, LBDC, L CDA.
Chapter Summary 359
in Exercises 2734, determine if the given statement is true or false. (An
ifthen statement is false if there are one or more instances in which the
ifpart is Lrue and the thenpart false.)
27. If a line is perpendicular to two distinct intersecting lines at their point
of intersection, then it is perpendicular to the plane that contains the
intersecting lines.
28* If the intersection of a plane and a dihedral angle is an angle., then that
angle is a plane angle of the dihedral angle.
29. If Z and n arc distinct intersecting lines and if I is parallel to plane « t
then n is not parallel to plane a.
30. If T and n are distinct parallel lines intersecting (hut not contained in)
two distinct parallel planes, then the planes cut off segments of equal
length on the two lines.
3t< If two planes are perpendicular to the same line, they arc parallel.
32. If two Kltet are parallel to the same plane, they are parallel.
33* Given a plane a, the set of all points each of which is at a distance of
5 from a is the union of two planes each parallel to a.
34. If a is a plane and F is a point, there is one and only one plane contain
ing V ami parallel to a.
CHAPTER SUMMARY
We l>egan the chapter with some suggestions for drawing figures to
represent geometrical figures that do not lie in a single plane. We used fig
ures throughout to help you visualize the relationships of lines and planes
in space.
Definitions of the following expressions were included.
A LINK AND A PLANE .ARE PERPENDICULAR
THE PERPENDICULAR BISECTING PLANE OF A SEGMENT
PARALLEL PLANES
A LINE AND A PLANE ARE PARALLEL
A PLANE ANGLE OF A DIHEDRAL ANGLE
THE MEASURE OF A DIHEDRAL ANGLE
A RIGHT DIHEDRAL ANGLE
PERPENDICULAR PLANES
A SEGMENT (RAY) PERPENDICULAR TO A PLANE
THE PROJECTION OF A SET ON A PLANE
THE DISTANCE BETWEEN A POINT AND A PLANE
THE DISTANCE BETWEEN TWO PARALLEL PLANES
There are 27 theorems in this chapter. It is suggested that you write
out these theorems and draw appropriate figures for them.
360 Perpendicularity and Parallelism In Space Chapter 8
REVIEW EXERCISES
■ In Exercises 115, copy and complete the given statement to obtain a the
orem of this chapter.
1. If a line is perpendicular to each of two distinct intersecting lines at
their point of intersection, tl icn 3
2. Given a line and a point, there is a unique plane perpendicular [T].
3. The perpendicular bisecting plane of a segment is the set of all points
each oi' which is TJ,
4. If two lines are perpendicular to the same plane, they are ]TJ,
5. If one of two distinct parallel lines is perpendicular to a plane, then \T\.
6. Given a plane and a point, there is a unique line containing the given
point and T.
7. If a plane intersects one of two distinct parallel lines but does not con
tain ft, then it \TJ and does not contain it.
8. If a plane intersects each of Lwo distinct, parallel planes, the intersec
tions arc [7] lines.
9. If a, /?, y are three distinct planes such that a is parallel to /? and such
that n intersects y t then [?]
10. If a line intersects one of two distinct parallel planes in a single point,
thenQ].
11. If a line is parallel to one of two distinct parallel planes, then \T}.
12. Given a plane and a point, there is a unique plane that contains the
given point and [?].
13. Any two plane angles of a dihedral angle are [7}*
14. If a line is perpendicular to a plane, then every plane containing that
line Is Q].
15. If two distinct intersecting planes are perpendicular to a third plane,
then their line of intersection is QJ
■ In Exercises 1623, some lines or planes are described. In each exercise,
state whether or not they must be parallel to each other,
16. Lines through a given point parallel to a given line.
17. Lines perpendicular to a given plane.
IS. Lines perpendicular to a given line.
19. Lines parallel to a given plane.
20. Planes parallel to a given plane.
21. Planes perpendicular to a given plane.
22. Planes perpendicular to a given line.
23. Planes parallel to a given lino.
Review Exercises 361
In Exercises 2430, write a plan for a proof of the given statement.
24. If TF is the projection of A£ on a plane, then A'B' < AB.
25. If A'B' is the projection of AB on a plane a and if A'B' — AB, then
AB  a.
26. If a and B are distinct parallel planes, if parallel lines m and n are in /?,
and if lines m' and n f are the projections of m and n, respectively, on « T
then m' is parallel to n'„
27. If « and /J are distinct parallel planes, if $ and f are parallel and con
gruent segments in t and if s' and f arc the projections of * and t, re
spectively, on a, then $' and f are congruent.
28. If A$ I ifi?', AC 1 1 A 7 ?, and if A, B, C are noneollincar points, then
plane ABC and plane A'B'C are parallel.
29. If a and /? are distinct parallel planes and if I and m are distinct parallel
lines that intersect ft in points Li and Mi, respectively, and /? in points
J2 and I4& respectively, then LiAfiJUgLi is a parallelogram.
30. If every plane containing a given line is perpendicular to a given plane,
then the given line is perpendicular to the given plane.
er
The Public Archives of Canada
Area
and the
Pythagorean Theorem
9.1 INTRODUCTION
You are familiar with the idea of area and have computed the areas
enclosed by figures such as triangles, sqtiaras, and circles. As you know,
areas are usually computed using numbers which arc lengths or dis
tances. You also know that a distance function is based on some seg
ment as the unit of distance. In informal geometry, we usually combine
a number and a word in expressing a distance, for example, 4.5 ft. In
formal geometry, we frerjuendy use the number by itself, omitting the
unit when the distance function is understood. Tn our formal develop
ment of the area concept wc shall suppose that a distance function is
given. Then there will be just one area function
In this chapter we adopt some postulates for area based on our ex
periences in computing areas and on our ideas about areas in informal
geometry. These postulates include the formula for computing the area
enclosed by a rectangle as well as statements of general properties that
are useful in proving theorems about area. We also develop formulas
for computing the areas enclosed by figures such as triangles and trape
zoids, and sve use area as a tool in proving the Pythagorean Theorem.
364 Area and the Pythagorean Theorem
Chapter 9
9,2 AREA IDEAS
Let us suppose that a unit segment for distance measure is given
and that a square of side length I is given.
We call this square the unit square. (See Fig
ure 91.) Just as distances and lengths are
numbers associated with sets of points (for
example, the length AH is associated with
AH), areas are numbers associated with sets
of points. The distance function matches a
positive number with each segment. Simi
larly, the area function matches a positive
number with each rectangle and also with
many other simple figures,
We shall agree that the area of the unit square is J , In practical ap
plications there is a system of areas for each system of distances. If the
side length of the unit square is 1 ft., then the area of the unit square is
1 sq. ft. If the side length of the unit square is 1 cm,, then the area of
the unit square is 1 sq, cm. (See Figure 92.)
F%v«Bi
1 aq. Lin.
Figure 92
1 so. in,
When we speak of the area of the unit square as 1 , we are thinking
of 1 as the measure of the set that includes afl of the points in the plane
of the square that are inside of or on the
square. (See Figure 93.) Strictly speaking,
it would be better to speak of the area of die
unit square region. We shall follow custom,
however, and call it the area of the unit
square. Similarly, we talk about areas of rec
tangles, triangles, and circles when we really
mean the areas of the regions which are made
up of these figures including their interiors.
(See Figure 94.)
When we say the "area of a triangle," for example, we mean the
"area of the triangular region."
Fljjure 93
9.2 Area Ideas
365
Definition 9.1 A polygonal region is a triangular region, or
it is the union of a finite numher (two or more) of triangular
regions such that the intersection of every two of them is
the null set, or a vertex of each of them, or a side of each of
them.
Polygonal
region
Union of
triangular
run ions
Figure 94
You may already know how
to compute the area of a rec
tangle. If its sides are of length
a and b, its area is ah, (See Fig
ure 95,) In Section 9.3 we
adopt this formula for the area
of a rectangle as a postulate.
Let us see if it is really a reason
able and basic assumption to
ure95
Figure 96 suggests our unit square and a rectangle with sides of
lengths 3 and 4. The rectangular region is made up of 3 * 4 = 12 square
regions each of the same Size as the unit square region. So the area of
the rectangle should be 3 • 4, or 12. Similarly if a and b are any two
natural numbers (that is, positive whole numbers), then the area of a
rectangle which is a by b should be ab.
□
Figure 86
I
366 Area and the Pythagorean Theorem
Chapter 9
Figure 97 suggests our unit square and another square with sides
of length ^, Let x denote the area of the square with sides of length L
Since it lakes 9 squares of this size (3 rows with
3 squares in each row as suggested in the figure)
to fill up the unit square, you can see that the
area of the unit square should be 9 times the
area x of the ^ by y square. Therefore
9x= 1
□
* — 1
x $.
l
Figure 17
It appears, then, that the area formula,
S = ah,
is a reasonable formula for our ^ by j square. Taking as and h = k
we have
1
the area of a ^ by ^ square. Similarly, if n is any natural number, then
area of a — by — square is — • — , or =r.
n
Figure 98 shows a rectan
gle which is ^ by . It is easy to
see from die figure that the area
of the rectangle should be 2 • 5,
or 1 0, times the area of a } by 4
square; hence
Therefore the formula tS as ah is appropriate for a rectangle which is
 by , Similarly, if a and b arc any two positive rational number*, then
the area S of a rectangle which is a by b should be given by the formula
S = ah. For example, if
_
i
Ekpre 98
then
6 = ,
I, _ 3 _ 24
" — 5 ~ 4I>*
and if R is a rectangle which is a by £>, its area S is the sum of the areas
75 • 24 = 1800 Utde squares,
each of which is 4 ' by £$, Hence
Since lgo = =Jf 4=(lM).
we see that the formula S = ah holds, in this case.
9,2 Area Ideas
Let R be a rectangle which is a by b and suppose that either a or b t
perhaps both, is an irrational number. Then, using the idea that an
irrational number can be squeezed between two rational numbers that
arc as close together as you desire, we can make it seem reasonable
that the area S of R is again given by the formula S = ah, Figure 99
shows a rectangle R which is 1 by yf%, (Recall that \/2 is not rational
and that i/2 can be represented by the nonrepeating decimal
1.41421 . . . .where the three dots indicate an infinite sequence of
digits.) Thus
L4< \/2<1.5>
where 1 .4 and 1.5 are rational numbers, Now a rectangle 1 by 1.4 has
area 1.4, a rectangle 1 by L5 has area 1.5, and therefore the area S
of the 1 by \*2 rectangle R is somewhere l>ctween 1.4 and 1 .5, that \s t
1.4 < S < 1.5.
367
Figure 99
Continuing in this way* we can show that S is between 1.41 and
1.42, between 1.414 and 1.415, and so on. If you consider the
sequence of areas of rectangles 1.4, 1.41, 1.414, . . . , you see that the
sequence is increasing and the rational numbers in the sequence are
getting closer and closer to \/2 through numbers that are less than \/2,
Similarly, if you consider the sequence of areas of rectangles 1.5, 1.42,
1*415, . * . , you see that the sequence is decreasing and the rational
numbers in the sequence are getting closer and closer to y/2, through
muni tens that are greater than y/2. It seems reasonable, then, to con
clude that the area S of the 1 by \/2 rectangle R is given by
Ssl'yfs y^.
The foregoing discussion may be summarized as follows. If the
sides of a rectangle are rational numbers, the rectangular region can
be either subdivided into a finite number of unit square regions and the
area obtained by counting, or subdivided into a finite number of square
regions each — by — for some integer n > 2 and then counting and
368 Area and the Pythagorean Theorem Chapter 9
multiplying byi In both cases the result S is the same as that obtained
by using the formula 5 = ah, where a and h are the lengths of two
adjacent sides of die rectangle. If the side lengths are not rational, then
we cannot subdivide the rectangular region into square regions with a
rational side length and find the area by counting, or by counting and
multiplying by ^ for some natural number a. We can, however,
n 2
approximate the side lengths using rational numbers and consider the
limit of the sequence of areas of the rectangles whose sides have
lengths that are rational numbers. It can be proved that this limit exists
and is equal to the product ah* In this book we adopt the formula S=al>
for the area of a rectangle as a postulate.
The ideas used to develop the formula S" = ah for the area of a
rectangle are, of course, the same ideas with which you are familiar
from your past experiences in informal geometry. They include the
following: (1) every rectangle has an area, (2) congruent rectangles
have the same area> and (3) area is additive in die sense that the area
of a figure is the sum of the areas of the parts of the figure. Similar
ideas may be used in developing area formulas for triangles, parallelo
grams, trapezoids, and so on. In Section 9.3 we state carefully our basic
ideas about area We call these statements die Area Postulates and use
them to develop several area formulas.
9.3 AREA POSTULATES
Our formal development of area in this chapter is limited to areas
of polygons and is based on the following postulates.
POSTULATE 27 (Area Existence Postulate) Every polygon
has an area and that area is a (unique) positive nuxnlier.
POSTULATE 81 (Rectangle Area Postulate) If two adjacent
sides of a rectangle are of lengths a and b t then the area S of the rectan
gle is given by the formula S = ah.
POSTULATE 29 (Area Congruence Postulate) Congruent poly
gons have equal areas,
POSTULATE 30 (Area Addition Postulate) If a polygonal re
gion is partitioned into a finite number of polygonal subregions by a
finite number of segments (called boundary segments) such that no
two subregions have points in common except for points on the bound
ary segments, then the area of the region is the sum of the areas of the
9.3 Ares Postulates
169
Notation, If 3 is a figure, we frequently use ff  to denote the area of
?. For example, &ABC\ denotes the area of A ABC. The area of quad
rilateral A BCD may he denoted by  quadrilateral ABCD\ or by \ABCD\
if it is clear that ABCD is a quadrilateral.
Example J Figure 910 suggests a polygonal region ABCDEFG, Let
us call this region (P. Segments CO and CP partition, or separate, (P
into three subregions (Pi, (P2, (P3, where (Pi, is the union of quadrilat
eral ABCC and its interior, <S>z is the union of A CFG and its interior,
and (P3 is the union of quadrilateral CDEF and its interior. According
to Postulate 30 } the Area Addition Postulate, the area of (P is the sum of
the areas of 6*1, (P & and (P3, that is,
<P = ((P^ + <Pa + (P 3 .
H C
Example 2 Figure 9 1 1 suggests a figure consisting of a square A BCD
with its diagonals, AC and DB t and an adjoining isosceles right triangle,
ABFC t with legs of length I, Let S denote the area of &BFC, S' the
area of square ABCD, and S" the area of pentagon ABFCD. FindS,
S\ S" using the Area Postulates.
Figure Ull
Solution: It is easy to show that ABFC = A BEC SO that the triangles
have equal areas and thai BFCE is a rectangle, actually a unit square.
Then
1 = \BFCE\
= \&BFC\ + \ABEC\
= J auk; + \ABFC\
by the Rectangle Area Postulate
by the Area Addition Postulate
by the Area Congruence Postulate
370 Area and the Pythagorean Theorem Chapter 9
Therefore 2S = 1 and S = j. Since square ABCD is partitioned by its
diagonals into four triangles, each congruent to A BFC, it follows from
the Area Addition and Congruence Postulates that S' = 4S and hence
that
S' = 4 • j = 2.
From the Area Addition Postulate it follows that
S" = S' + S = 2J.
Another way to find S' is as follows. Since ABbXl = ABEC, we
have
BE = EC=l and mlBEC = 9Q.
Then it follows from the Pythagorean Theorem, which is proved in
Section 9.5, that
(BC)2 = 12 + 12 = 2 and BC = \M.
Finally, it follows from the rectangle area formula that
S' = y/2 • \/£ = 2
EXERCISES U
■ In Exercises 18, given the lengtta a and 6 of two adjacent sides of a rectan
gle, find the area of the rectangle.
L a = 2,5, b = 3.3 5« = £ b = 4
2.a = % t b = i 0, a = 7, b = 3yl"
3.as^s 7, a = 4t/2, 6 = syi
4 a = 2,32, 6 = 1.77 8, a= Vo y b = 5^
9. (An informal geometry exercise.) The length of a side of a square is
10 yd. What is the area of the square in square yards? In square feet?
10, (An informal geometry exercise.) ]f the dimensions of a rectangular
field are 40 rods by 80 rods* how many acres of land does it contain?
(An acre contains 43,360 sq. ft. and 1 rod is 16 ft. long.)
1L (An informal geometry exervim.) How many feet of fence would it
take to enclose the field in Exercise 10?
12. (An informal geometry exercise.) The dimensions of a football field
(including the end zones) are 53} yd, wide by 120 yd long. Would an
acre of land be sufficient on which to lay out a football field? (See Ex
ercise 10,)
13. If the perimeter of a square is 4H, what is its area?
14* If the perimeter of a square is 64 \/2. what is its area?
9,3 Area Postulates
371
15 The length of one side of a rectangle is 3 1 hues the length of an adjacent
side. If the perimeter of the rectangle is 5fi, what is its area?
16. The length of one side of u rectangle is 5 more than twiee the length of
an adjacent side. If the perimeter of the rectangle is 70, what is its area?
17. challenck pkoblem. The length of a rectangle is 4 more than twice
its width. If the area of the rectangle is 48, what is its perimeter?
IS. The width of a rectangle is  of its length. If the area of the rectangle
is 360, find its length and width.
U). (An informal geometry exercise.) How many squares 1 in. on a side
are contained in a square 1 yd. on a side?
20. (An informal geometry exercise.) How many 9 in. X 9 in. tiles would
it take to cover a 12 \ ft. X 9 ft. rectangular floor if wc assume that there
is no waste in cutting the tile?
21. If S is the area of an x by x square and S' is the area of a 2.v by 2.v square,
prove that S" = 43,
22. If S is the area of a right triangle with legs of lengths a and b t and 5' is
the area of an a by b rectangle, prove that S* = 2S.
23. If S is the area of a right triangle with legs of lengths a and b t and 5' is
the area of a right triangle with legs of lengths 2a and 2b, prove S' = 4S.
24. The figure shows two coplanar equi
lateral triangles, AABC and AACP,
with AC = 1 . The feet of the perpen
diculars from A and D to EC are E
and F, respectively. Prove that ABCD
is a parallelogram and that AEFD is a
rectangle.
25. Complete the proof of the following theorem.
THEOREM The area of a right triangle is onehalf the product of
the lengths of its legs.
restatkmkst:
Given: A right triangle A ABC with
the right angle at C, AC=b 7
and DC = a.
_ i
Pram. AAJ9C =
Proof: Let C be the point of intersection of the line through A and
parallel to OB and the line through B and parallel to XU, (See the figure,)
Why must these two lines intersect? YVhy is quadrilateral ACBC a
rectangle?
Complete the proof by showing that AABC ^ ABAC and that
 A ABC = \ab.
372 Ar«i And the Pythagorean Theorem
Chapter 9
9.4 AREA FORMULAS
Figure 912 shows a parallelogram ABCD with points £ and F the
feet of the perpendiculars from D and C, respectively, to line AB.
Figure 912
The following definition has two parts. In (1) we define base and
altitude, thought of as segments. In (2) wc define base arid altitude.
thought of as numbers (lengths of segments, or distances). (Compare
the corresponding definitions for triangles on page 265.)
Definition 9.2
1 . Any side of a parallelogram is a base of that parallelogram.
Given a base of a parallelogram, a segment perpendicular
to that base, with one endpoint on the line containing the
base and the other endpoint on the line containing the
opposite side, is an altitude of the parallelogram corre
sponding to that base,
2, The length of any side of a parallelogram is a base of that
parallelogram. The distance between the parallel lines con
taining that side and the side that is opposite to it is the cor
responding altitude. Of height
In Figure 912 we call h the altitude, or height, corresponding to
the base AB. The number h is the distance between the parallel lines
A3 and DC. Similarly, the distance between the parallel lines AD and
BC is the altitude which corresponds to the base AD,
Let S denote the area of parallelogram ABCD in Figure 912, let
AB = ft, and let CF s ft, Wc proceed to prove that the area S of the
parallelogram is bh.
THEOREM 9.1 If h is a base of a parallelogram and if h is the
corresponding height, then the area S of the parallelogram is given
by the formula $ = bh.
9.4 Area Formulas
373
talement
L £DAE=& /CBF
2. lAEDj= IBFC
3. Z5E ss CF
4. AAED m ABFC
5. \AAED\ = \ABF€\
a \AFCD = S + ABrcj
7. \AFCD\ = S+ AA£D'
8. S= AFCD  AAED
9. £FC£> + AAED\ =
\AFCD\
in. OC = b
1L EFCD = fc/i
12. fefi + AAFD = \AFCD\
13. /?/i = AFCD  \AAED\
14. S = Mi
Reason
1. Why?
2. Why?
3. Why?
4. S.A.A, Theorem
5. Why?
8. Why?
7. Why?
8. Addition Property of Equal
ity
9. Why?
10. Why?
11. Why?
12. Steps 9, 11; substitution
13. Why?
14 Steps 8, 13; substitution
In the remainder of this section we prove two more theorems re
area formulas.
THEOREM 9.2 If b is a base of a triangle and if h is the corre
sponding altitude, then die area S of the triangle is given by the
formula S = ^bh.
Proof: (See Figure 9 13 . ) Let A A3 C be given . Let 55 be the ray with
endpoint B and parallel to AC. Let CE be the ray with endpoint C and
parallel to AB. Let F be the point of intersection of bB and CE. Then
ABFC is a parallelogram and A ABC SB A FOB.
Figure 913
374 Area and the Pythagorean Theorem Chapter 9
Now b m A3 and h is the distance from C to A$> and so /i is also the
distance between CE and S3, It follows from Theorem 9.1 that
ABFC\ = bh> But
I A ABC] = \AFCB\ (Why?)
and
AABC + AFCB = \ABFC\ (Why?).
Therefore
S + S = 6Ji, 2S = &/», and S = Jbfc.
THEOREM 9.3 If 6i and b 2 are the lengths of the parallel sides
of a trapezoid and if h is the distance between the lines that con
tain these parallel sides, then the area S of the trapezoid is given by
die formula
S s 4(&i + b*)h.
Proof: Let trapezoid ABCD t with parallel sides AH and CD, be given
as in Figure 914.
h
/
A
Figure 914
______ — . ,
The lengths of these sides, or bases, are b\ and h% % and the distance
between AB and CD is h. The distance to is called the height. Let E
be the point on opp BA such that BE = Ify. Let F be the point on
opp CD such that CF m b%. Then AEFD is a parallelogram (Why?) and
its area is (bj + & 2 )n. We shall show that
A BCD «— * FCBF
is a congruence, Then it will follow that
ABCD + \FCBE\ = \AEFD\
S + S = (hi + &_)fc
S = •£(&! 4 &)fc.
9,4 Area For mulit 375
To see that ABCD «— * FCBE is a congruence, note that
(1) LBAB = lEFC (5) XB E PC
(2) ZABC a ZFCB (6) B'C <= GB
(3) IBCD=* ICBE (7)C35 = B£
(4) LCDA = ABEF (8) DA sj EF
This completes the proof.
There are area formulas for other special polygons, but we shall not
develop them here. In Section 9.5 we use area formulas us tools in prov
ing the Pythagorean Theorem.
EXERCISES 9.4
I* If the area of a triangle is 40 and a base is S.. find the corresponding
altitude,
2. If an altitude of a triangle is 12 and the area of the triangle is 62, find the
base corresponding to the given altitude.
3. The lengths of the legs ol a right triangle are 17 and 9. Find the area of
the triangle.
4* Find the area of an isosceles right triangle if the length of one of its legs
is 15,
5. Find the area of a rhombus if its height is 7j and the length of one of its
sides is 15^.
■ Exercises 68 are informal geometry exercises.
6. In the figure below, lines m and n are parallel and the distance between
them is 5 ft. P and Q are points on m, and R and S are points on n such
that FQ = RS = 7 ft. If the distance from P to R k 10 miles, find
PQRS.
P Q
4 — • ♦ 7/ btn
8
M
7. In the situation of Exercise 6, find FQHS if PR = 1 mile.
& In the situation of Exerd.se 6, find \PQRS\ if PR = 10 ft,
9* If the area of a triangle is S and its altitude is h, express the base b in
terms of S and h.
10. The area of a parallelogram is 50.4. Find its base if the altitude is 4.2.
376 Area and the Pythagorean Theorem Chapter 9
11. Find the area of a trapezoid if its height is 7 and the lengths of the par
allel bases are 9 and 15*
12. Use the formula for the area of a trapezoid and express h in terms of
5, hi, and b&
13. If the area of a trapezoid ( s 128 and the lengths of the parallel bases are
7 and 9, find its height,
14. The area of a trapezoid is 147 and the lengLh of one of i ts parallel bases
is 18. If the height is 7. find line length of the other base.
15. ABCD is a parallelogram with AB = 25. If \ABCD\ = 375 and P is a
point such that CPD, find  AAPB.
Ifl. An informal geometry exercise.) A plot of land in the shape of a
trapezoid has bases of lengths 121 yd. and 242 yd. If the distance be
tween the bases is 300 yd., how many acres of land are contained in the
plot? (One acre contains 43,560 sq. ft)
17. Two triangles have the same base and their areas differ by 42, If the
altitude of the larger triangle is 6 more than the altitude of the smaller
one, find the length of the common base.
18. Id the figure, ABCD is a trapezoid with parallel liases AB and CD. If
AB m fej, CD = h z , and the altitude is h, draw ED and use the Area
Postulates and the area formula for a triangle to prove that
\ABCD\ = ^(fcx + /ia).
19. In the figure, AM is a median of AABC. Prove that
lAAJMfj = AACMj.
Is AABM r A ACM in every case? Iu any case?
The area of a square is equal to the area of a parallelogram . If the base
of the parallelogram is 32 and its height is 18, find the perimeter of the
square.
9.4 Ar»a Formulas 377
In each of Exercises 2129, find, on the basis of the given data, the area or
areas indicated,
21, X3 I Be* BC = 8, 23, ABCD is a square,
DE = 6. f AAfiCJ = [T]. E is the midpoint of HC
The perimeter of ABCD is 48,
A D IAACE =m.
22. DGF r DF = 6, DE a 6.5,
GE = 5. ADEF] = {T\.
26. ABCD is a parallelogram, E is
the midpoint of C~D r and
A£CDj =33.
JAA£D] =[?]
jAA£E =0]
ASCE = fl
23, AH = CD = 8, BC = DA = 6,
DE b 5. AABCD\ = JTj.
27. A£CD is a square; K, F, G, //
are midpoints of the sides;
jABCD, = 196, \EFGH\ = [?].
ABCD is a parallelogram, EF ±
M, £F =3, BC= 8, CD = 4.
AA/*CD = [2].
378 Area and the Pythagorean Theorem
Chapter 9
2H. EFGH is a square of side length 6; ABCD is a square of sick length 12;
E, F s C, It are interior points of ABCD. The area of the shaded region
29. B and £ arc collinear with A and F ( K and
H are collinear with L and (». C and / are
collinear with B and /C, D and J are col
linear with E and H, AFCL is a rectangle
with area 46, BCDE b ;i iert,mgjB with
area 8, BCDE and jfKHJ are congruent
rcctangjes. ABCDEFGHIJKL\ = [J
fl
J AT
J H
30. Using the Area Postulates and the theorem of Exercise 25 in Exercises
9.3, derive the tri angle area formula for any triangle.
31. (See Figure 912.) Let b\ = AD and let A, be the distance between
AD and BC. Is £?i/»i = hh? Justify your answer.
32. Would any of the statements in the proof of Theorem 9,1 be different
if point £ were between A and B? Draw a figure for this case.
33. See the proof of Theorem 9.2. Prove that Q? and BD are not parallel
lines.
34 In the proof of Theorem 9.2, prove that A ABC = AFCB.
35. Recall that a parallelogram is a special trapezoid. Show that the formula
for the area of a trapezoid simplifies to the formula for the area of a
parallelogram if the trapezoid is a parallelogram.
36. A triangle might be thought of as a quadrilateral in which one of the
bases has shrunk to a point. What number does j[hi + b%)h approach
if h t and h are fixed and h% gets closer and closer to 0?
37. challenge problem. Prove that the area of the triangle determined
by the midpoints of the sides of a given triangle is onefourth the area
of the given triangle,
9.5 Pythagorean Theorem 379
9.5 PYTHAGOREAN THEOREM
In this section we state and prove the Pythagorean Theorem and
its converse. Perhaps no other theorem In mathematics has inspired
more mathematicians and nonmathematicians alike to find original
proofs as has this one. There are over 370 known proofs of this the
orem of which over 255 employ the use of areas. The first proof of this
theorem has been attributed by many historians to Pythagoras (582
501 Ji.c), a Greek mathematician and philosopher, although the prac
tical application of the theorem was known many years before his
time. It is not known which proof Pythagoras gave, but in the Exercises
at the end of this section is an outline of a proof which appeared altoul
300 b.c. in the first of the thirteen books of Euclid's Elements.
Perhaps the most unique proof of the Pythagorean Theorem was
devised by a 16yearold schoolgirl, Miss Ann Condi t, of South Bend,
Indiana, in 1938. The proof devised by Miss Condit and an original
proof devised in 1939 by Mr. Joseph Zelson, an 18yearold junior in
West Philadelphia, Pennsylvania High School, show that high school
students are capable of original deductive reasoning. For a list of proofs
of the Pythagorean Theorem, including the proofs by Miss Condit and
Mr. Zelson, see The Pythagorean Proposition by Elisha Scott Loomis,
published by the National Council of Teachers of Mathematics, 1968.
THEOREM 9.4 (The Pythagorean Theorem) If a, b, c are the
lengths of the sides of a right triangle A ABC, with c = AB t the
length of the hypotenuse, then
a 2 + &>=<&
Let BC = a, CA = h. Let PQRS be a square of side length
a + b as in Figure 915. The points T, (/, V, W are interior points of
PQ, QR* R§, SP, respectively, such that FT = QU = RV = SW = a.
T b Q
380 Area and the Pythagorean Theorem
Chapter 9
Then TQ = UR = VS = W? = h, and ATQU. AURV, AVSW,
A WFT are four congruent right triangles, each with area ^ih. Hence
TUVW is an equilateral quadrilateral. Since die two acute angles in a
right triangle arc complementary angles, it follows that
nuLWTU =180 nUPTW  nUUTQ
 mlPTWmlPWT
 (mlPTW+nUFWT)
z 180  90 = 90.
By a similar argument it can be shown that the other three angles of
TUVW toe right angles. Therefore TUVW is a square. Let x be its side
length. Then
(a + fr) 2 = \PQRS
(a + fc) 2 = 4 AWPTj + \TUVW\
{a + bf = 4 * \ab + x 2
(a + &) 2 = 2al> + x 2
a 2 + 2d/> + fc 2 = 2ab + x 2
a 2 + fr 2 = x 2 .
It follows from the S.A.S. Congmence Postulate that A WPT =
AABG Therefore
x = c and a 2 + fe* = e 2 .
THEOREM 9.5 (Convene of The Pythagorean Theorem) If a,
b, c are the lengths of the sides of a triangle and if a 2 + b 2 = c 2 ,
then the triangle is a right triangle and the right angle is opposite
the side of length c.
Proof: Let there be given AARC with AB  c, BC = o, CA = b,
and a 2 + fc* = c; 2 . Let AA'B'C l>e a right triangle with B'C = a,
CA' = b, A'B' = x, and the right angle at C" as shown in Figure 916.
C o
Figure SUA
9.5 Pythagorean Theorem 381
Then
a2 + fez _ *2. Why?
By hypothesis, a 2 4 h 2 = c 2 > By substitution, x 2 = c 2 and, since x >
and c > 0, it follows that xc. Therefore A ABC s AA'B'C'by the
S.S.S. Postulate and LC^LC It follows that ZC is a right angle
and that A ABC is a right triangle.
When solving problems by use of the Pythagorean Theorem, it
is often necessary to find the square mot of a number that is not a per
fect square. Recall that a rational number is a perfect square if it can
be expressed as the square of a rational number. Thus 16 and 2 ,f are
perfect squares because
16 = (4)2 and V = ($)*,
whereas 32 and ^ arc not perfect squares. If we arc required to find
the square root of a number like 32 or ^ t we can approximate the
square root by means of a rational number or we can leave our answer
in what is called simplest radical form, If % = y/A, where A is a posi
tive rational number which is not a perfect square, we say that x is in
simplest radical form when it is expressed as B>/C t where B is a ra
tional number and C is a positive integer which contains no factor
(other than 1) that is a perfect square.
We make use of the following theorems from algebra when putting
radicals in simplest form,
1. If a and b are positive numbers, then
■\/ab — Vff * \/5*
2. If a and b are positive numbers, then
ft£_ Va
V b yE'
Example I Put (1) y/3£, (2) >/¥"' and (3) \/105 hi simplest radical
form.
i
Solution: 1. V32 = \/W 2 = y/TB • y/2 m 4 \/l"
" / 8 " JT' 2 = X /I6 " 4 "
3. \/I06 is already in simplest radical form. Why:
382 Area and the Pythagorean Theortm
Chapter 9
EXERCISES 9.5
In Exercises 115, put the indicated radical in simplest radical form.
1 V% 9, \/2M
& \/!8 10. \/L35
3. v§7
4. y^O
5. v^S
6. yfffi
7. y300
11. vf
13. v*?
x/ 12
~rr
16. If ci 2 + b 2 = r; 2 , solve for a in terms of fc and c,
17. If o 2 + LP = c 2 , solve for b in terms of a andc.
IS. If « 2 + b 2 = e 2 , solve for r; in terms of a and &,
19. If the length of a diagonal ttf a square is 15, find the area of the square.
{Hint: Use the Pythagorean Theorem,}
20. In the figure, ABCD is a parallelogram, BE ± AB at E t AD  15,
CD = 20, and A£  9, Find the area olABCD.
<> !■:
20
In Exercises 2128, A ABC is a right triangle with right angle at C, a = BC,
b = CA, c = AB, In each exercise, two of the three numbers a, b r c are
given. Find the third one. Express each answer in decimal form correct to
two decimal places.
21. a = 14 h = 1 jO
22 a = 1.0, b = 2.0
23. a= 10.0 f fo= 10.0
24. a = 3.0, 6 = 4.0
25. a = 9,0, b = 40.0
26. c = 3.1, fo = 5.2
27. a = 10.0, fc = 9.0
28. c ^ 10.0, a = 7,0
9.5 Pythagorean Theorem 383
In iLxereises 2936, A ABC is a right triangle with right angle at C, a = BC t
b = CA, G ss AM, In each exercise, two of the three numbers a, b, c are
given. Find the third one, Kxpress each answer exactly and in simplest form,
using a radical if needed.
29. a  7, b = 40
30. a = 39, c = m
31. a = \/2, h = vT
32. = v/5I. fe = \ ^
33. a = J, fc = i
34. o = 3x, £? = 4x (Find c in terms of *,)
35. a = x + gft & = x — y (Find c in terms of x and y.)
3ft. a = u 2 + C^ f a = a 3 — u 2 (Find b in terms of u and u.)
37. See the proof of Theorem 9.4, From which Triangle Congruence Pos
tulate docs the conclusion that ATQV s A URV follow?
38. See the proof of Theorem 9.4. From which Area Postulate does it fol
low that the area of the large square is the sum of the areas of the four
triangles and the smaller square?
39. Sec the proof of Theorem 9.4. From which Area Postulate does it fol
low that the four triangles have equal areas?
40. In the proof of Theorem 9.4, which Area Postulate supports the con
clusion that the area of TUVW is x 2 ?
41. Let AABC with BC = a, AC == b, AB = v, c>b,c>a, and
c 1 > « 2 4" b 2 be given. Answer the "Whys?" in the following proof
that L C is an obtuse angle
Proof; Let AA'B'C He a right triangle with B'C = a, CA' = b,
A'B' = x, and the right angle at C as shown in the figure. Then
a* + t>i s s*
for AA'B'C, Why? By hypothesis, a 2 + ^ < # for A ABC. By sub
stitution, x 2 < c 2 and, since x > and o > 0, it follows that x < c,
Therefore IC> LC Why? Since mZC' = 90, it follows that
m£C> 90. Theu L C is an obtuse angle. Why?
304 Area and the Pythagorean Theorem
Chapter 9
42. Let A ABC Willi BC  a t AC = b, AB = c, c > b, c > a, and c* <
a 2 + fe 2 be given Prove that / G is an acute angle. (See Kxercise 41)
43. Combine Theorem 9.5 and the statement* proved in Exercises 41 and
42 into a single theorem.
44. chaijlenge problem. The Pythagorean Theorem may be stated as
follows.
In a right triangle the area of the square on the hypotenuse
is equal to the sum of the areas of the squares on the legs,
iiestatement! A ABC is a right triangle with the right angle atC,
BC = a, AC = b, AB a c, squares ABDE, BCLK, and AIGC as
shown in Figure 9 17a. Prove Uiat
t ABDE\ = \BCLK\ + \AFCX:\
and hence that c a = a 2 + h*. Complete the following proof which up»
peared in Euclid's Elements around 300 b.cj.
Figure ft.17
Proof: Draw AK, CD, and €37 1 IM at M and intersecting A~B at X
as shown in Figure 91 7b.
Why?
(1) AABK ^ ADBC
(2) \AABK\ = \ADBC\
(3) The altitude from A to £3 of AABK is equal to BC.
(4) AABK =  * KB  »C = } • fl 2 Whv?
(5) JBCLiq =2* AABK  Wh^ff
(6) The attitLide from C to B3 of ADBC is equal to BN.
(7) 'ACBD, = lPBJJA'
Whv?
Why?
Why?
Chapter Summary 385
(8) BDMN\ = 2 *  ADBC\ Why?
(9) \BCLK[ = \BDMN\ Why?
(10) Draw BFand UE, ABAF = AEAC Why?
(11) \AFGQ =2*ABAf Why?
(12) A£»fiV = 2 •  A£Aq Why?
(13) fAFCCf = [A£.\/,Y Why?
(14) ABDE = BDMN\ + \AEX1X Why?
= \BCLK\ + AfGC Why?
(J 5) Therefore c 2 = « 2 + b 2 Why?
45 challexce problem. 11 your high school library contains the book
27*e Pythagorean Proposition, look up the proof devised by Miss Aim
Gondii: and present it to your class.
CHAPTER SUMMARY
In this chapter there a re four Area Postulates and several area formulas
stated ami proved as theorems. The last two theorems of the chapter are
the Pythagorean Theorem and its converse.
POSTULATES
Area Existence
Rectangle Area
Area Congruence
Area Addition
FORMULAS
Rectangle Area Formula S = ah
Parallelogram Area Formula S s hh
Triangle Area Formula S = \bh
Trapezoid Area Formula S = J(&i + b 2 )h
THE PYTHAGOREAN THEOREM. If a, b, c are the lengths of the
sides of a right triangle AA/JC, with c = AB> the length of the hypot
enuse > then a 2 + b 2 = c 8 .
You should be able to stale each of the postulates in your own words
and to explain the meaning of each of the formulas. You should be able to
use the postulates in developing the properties of area. You should under
stand the formulas so that yon can recognize a situation in which a formula
is applicable and then use it correctly.
386 Area and the Pythagorean Theorem Chapter 9
REVIEW EXERCISES
1. Find the area of a trapezoid with two parallel sides of lengths H and 13
if the distance between the lines containing those sides i 24.
2. If the hypotenuse of a right triangle is 50 in. long and one of the lcg?i is
40 im long, how long is the other leg?
3. The figure shows a right triangle A.ABC with right angle at A and with
D the foot of the perpendicular from A to EC. If BD  3, DC = 5{,
and Ai3 = 5, find \ A ABC
4. Find the area of a right triangle if the length of the hypotenuse is \/T
and the length of one log is y/5.
5. Find the base of a parallelogram if its area is 143 and its height is 7.
6. Find the altitude of a triangle corresponding to a base of length 12 if
the area of the triangle is 62,
7. If A ABC is an equilateral triangle of side length 6. find the area of the
triangle. (f/iitfcUsc the Pythagorean Theorem to find an altitude of the
triangle.)
8. See Exercise 7. Prove that the area S of an equilateral triangle of side
length f is given fay
■ J&.A
4
9. Use the result of Kxercise 8 to find the area of an equilateral triangle
of side length 12, Compare the area of this triangle with that of Kxercise
7. Is the ratio of the areas of the two triangles the same as the ratio of
the lengths of their sides? If not, how do the areas compare?
10. If ABCD is a parallelogram with m/ DAB — 45, DA a 12, and
DC = 21, find \ASCD\. (Mint: Draw ~DE _ AB at E, Which kind of
triangle is A DAE?)
In Exercises 1113, A ABC h given with a = EC, h = AC. and c = Afl
In each exercise the three numbers a, b, c are given. Is L C a right, an ob
tuse, or an acute angle? (See Exercise 43 of Exercises 9.5.)
11. a = ft6 = g,cs 12
12. a = 8, b = 15, c = 17
13. a = ll,b= 13, c = 17
Ftevtow Exercises 387
14 *!Tie area of a trapezoid is 164 and the distance between the parallel
bases is 10. If the length of one of the bases is 15, find the length of the
other base.
15. (An informal geometry exercise.) The dimensions of a shower stall
are 3 ft. by 3 ft. by 7 ft. How many 4 in. by 4 in. tiles are needed to
cover throe of the four rectangular walls if wc assume that there is no
waste in cutting the tile?
16. {An inform al geometry exercise, ) \ rectangt. Jar plot of land is 1 00 yd.
by 150 yd. A standard city lot for this particular plot of land is 75 ft.
by 150 ft. and sells for $3000 If a real estate agent can buy the plot of
land for $30,000, how much profit would he make if he divided it into
standard lots and sold all of thein? Disregard surveying and legal
expenses,
17. An isosceles triangle has two sides of length 13. If the altitude to the
base is 12, find the area of the triangle.
18. A rhombus has sides of length 10 and the measure of one of its angles
is 45. Find its area.
19. The length of the hypotenuse of a light triangle is 25 and the length of
one leg is 24,
(a) Find the length of the other leg.
(b) Find the area of the triangle.
(c) Find the altitude to the hypoteuusc.
20. A right triangle has legs of lengths 10 and 24 and hypotenuse of length
(a) Find the area of the triangle.
(b) Find the altitude to the hypotenuse.
21. The length of one base of a trapezoid is .5 more than twice the length of
the other base. If the area is 100 a net the altitude is 10, find the lengths
of the two bases.
22. In the figure, AD is an altitude of A ABC. If AD = 24, AB a 26, and
AC = 30, find BC and A ABC.
23. Recall that the diagonals of a rhombus are perpendicular to each other.
Use this fact to find the area of a rhombus whose diagonals are of lengths
10 and 8, How is the area of the rhombus related to the lengths of its
diagonals? State this relationship in the form of a theorem and prove it.
388 Area and the Pythagorean Theorem
24. In the figure, All _l W t BCD, AC = 10, BC
FindHDand jAAG'D,
Chapter 9
AD = 17.
25. In the figure, with right angles and congruent segments as marked, if
VA = 2andAB = 1, find VF.
26. Taking the area of a given unit square as I and using any of the Area
Postulates except the Rectangle Area Postulate, prove that the area of
a 4 by 5 rectangle is 20.
27. Taking the area of a given unit square as 1 and using any of the Area
Postulates except the Rectangle Area Postulate, prove that the area of
a i by i rectangle te £■
28. The figure suggests an alternate proof of the Pythagorean Theorem
which makes good use of the area formulas for a trapezoid and a tri
angle, Write out the main steps in this proof. (The proof suggested by
this figure is attributed to President Garfield and was discovered by
him around 1876.)
Review Exercises
381
29. challenge problem. The figure shows a triangle and its three medi
ans intersecting at the point G which is twothirds of the way from any
vertex to the midpoint of the opposite side. Prove that
30. challenge problem. Given the same situation as in Exercise 29,
prove that the six small triangles all have the same area.
Courtesy uf Leo GatieJU Gallery
Similarity
10,1 INTRODUCTION
Our experiences with objects of the same size and shape suggest
the concept of congruence in formal geometry. The idea of same shape
suggests the concept of similarity which you are ftbouA to study in this
chapter. Consider a picture and an enlargement of it. In the enlarge
ment each part has the same shape as it has in the original picture, but
not the same size. The picture and its enlargement are examples oi
similar figures.
Consider a floor plan for a building. The floor plan is a drawing
made up of segments labeled to show the lengths of the segments in the
actual building which the segments on the drawing represent. The
floor plan is similar to the actual floor of the building. Although the
plan and the floor do not have the same size, they surely have the same
shape. Each segment in this plan is much smaller than the segment it
represents in the building. We express this fact by saving that the plan
is a scale drawing of the floor. Each angle in the plan has the same size
as the angle it represents on the actual floor.
In drawing a floor plan, lengths arc reduced and angle measures are
preserved, The lengths of segments in a floor plan are proportional to
the lengths of the segments that they represent in the floor, and a state
ment of this fact is an example of a proportionality. In the next two
sections we define a proportionality and develop some of its properties.
392 Similarity
Chapter 10
In the remaining sections we develop the concept of similar figures.
The main theorems of this chapter have to do with triangle similarity.
The chapter includes a proof of the Pythagorean Theorem based on
similar triangles.
10.2 PROPORTIONALITY
Figure 101 shows two figures, one with side lengths labeled a, b,
c, d, e and the other with a\ b\ c\ d\ e\ Note that a = 2a\ b = W t
c = 2c\d = 2d", e = 2e\ and hence that
JL — JL — .£ , d__ e_ __ p
a' "" V " c' " d' " e f "
dmH
c3
• ?■ 9
c'*L5
d'mS
*' = as
a* 4
a'm2
e'«4
Figure 101
We can express the idea of these equations by saying that a, b, c, d r e
are proportional to a\ b\ c\ d\ e\ respectively.
Definition J 0.1 Let a onetoone correspondence between
the real numbers a, b, c, . . . and the real numbers a',h\ c f t
... in which a is matched with a', h is matched with b\ c is
matched with c\ and so on, be given. Then the numbers a, h,
c, ... are said to be proportional to the numbers a', b\ c', . . .
if there is a nonzero number k such that a = ka\ b = kh\
c = k& The number k is called the constant of
proportionality,
in Definition 10.1, why do you think k is required to be a nonzero
number?
10.2 Proportionality 393
Notation. We use = to mean "are proportional to." We use
(a, b> c s . . .) to represent an ordered set of numbers. Then
w
means that the numbers a,b,c,... are proportional to the numbers
d % h\d, t it being understood that a is matched with a\ h with f,
c with c', and so on. Note that the order of the numbers in the set
{a, h, c, , , ,) and in the set [a', b', c', . . .) is important only to the extent
that corresponding numbers must appear in the same order in the pro
portionality. Thus if we have
we could also write (h, c, a, . . .) = (bf, <f $ a\ . . .), and so on.
Example 1 (See Figure 101.) (4, 7, 3, 6, 8) = (2, 3.5, 1.5, 3, 4).
Example 2 (2, 3.5, 1.5, 3, 4) = (4, 7, 3, 6, 8).
Since
2 = ^4, 3.5 = 47, 15 = 4*3, 3 = ^6, 4 = ^8,
the constant of proportionality in Example 2 is j. What is the constant
of proportionality in Example 1?
Example 3 If (*, y t 9) = (7, 8, 6), find x and y.
Solution: There is a number k such that % = k ■ 7, y = k • 8,
9 = k • 6. From the last equation we have k = j, so
3 = 47 = 10.5 and y = §8=12.
Example 4 U the following statement true or false?
(5, 8, 10) = (8, 12.8, .16).
Solution; The statement is true if and only if there is a nonzero num
ber k such that
5=Jfc8
8 = it 12.8
10 m k* 16
From the first of these three equations, we see that k must l>c , or
0.625. We must now check to see if this k "works" in the other two
equations,
394 Similarity Chapter 10
Check. Is 6 = 0,625  12.8? Yes,
Is 10 =  • 16? Yes.
Therefore (5, 8, 10) = (8. 12.8, 16) is true.
Example 5 Ts the statement (4, 8, 10) = (6, 9, 16) true?
Solution:
If 4 = k • 6, then Jfc = f ,
Is 6 = j< 9? Yes,
Is 10=  16? No.
Is the statement (4, 8, 10) = (6, 9, 16) true? No.
EXERCISES 10.2
In Exercises 112, determine whether the given statement is true or false.
If it is true, find the constant of proportionality. The answer for Exercise I
has been given as a sample.
1. (1, 1, 5) = (3, 3, 15). TYue; k = \.
E. (3, 6, 9) = (2, 4, 6)
3. (5, 1) = (8. 2)
4. (8, 10, 15, 21) = (7, 1 1. 16, 22)
5. (5, 15, 25, 35) = (1, 3, 5, 7)
6. (2, 6, 7, 15) = (1, 3, 3.5, 7.5)
7. (1,3.1 +3) = {2, 6, 2 + 6)
8. (x, y. % + y) = (7x, ly t 7x + 7y)
9. (m, v,w) = {3«, 3t\ 3m>)
10, (0, 0, 0) = (5, 10, 15)
U. (5, 10, 15) = (0, 0, 0)
12. (0, 0, 3) = {0, 0, 4)
In Exercises 1 320, copy and complete the given statement so that it will
be true.
13. (8, 8) = (0,6)
14 mm, 100) = (5, 6, 10)
15. (5, 0,7) = ([7], 8, 10.5)
16. If (3, x) = (8, 10), then x = [7J.
17. If (7. 9) = (1x, 18), then a = [7J.
18. If (7. x) = (14, 21), then x = \t\>
19. If (5 t i) = {*, 125) and * > 0, then x = [7].
20. If (3, x) = (8, if) and i/ yt o t then {3, [3 ) = (*, y).
10.2 Proportionality 395
In Exercises 21 30, x is a positive number and (3. x) = (5, 15). In each ex
ercise, determine whether the given statement is true.
21. £ = ^
5 15
22. — = —
x 15
23. (3,*.3 + *) = {5, 15*20)
24. {x, 3  x) = (15. 10)
25. (*, 3) f (15, 5)
20. (3, 5) = (x, 15)
27. (3. 15) = fc 5)
28. (3 t 5) = {15\r)
29. 315 = x5
30.
3 5
x + 3 5+15
31. Does (fl, &, c) * — » (the,/) indicate the same onetoone corre
spondence as (a, c, b) * — * (d, /, *?)? As (c, a, fo) « — * (/, rf T e)? As
fcfea) *— * (/.«.d)?
32. If a, ft, e, <i are nonzero numbers such that (a, h) = [c, d) with pro
portionality constant 2 , is it true that (b,a) = {d, c) with proportionality
constant 2? Give your reasoning.
33. If a, b, c,d are nonzero numbers such that (a, b) = {c, d) with propor
tionality constant 2, is it true that (c, d) = (a f b) with proportionality
constant 2? Give your reasoning.
34. If a t b, c, d are nonzero numbers such that (a, b) = (c, </), prove that
(a, c) = {b, d).
35. If {a, b) = (c, if), prove that ad = he,
36. If ad = fee, prove that {a, b) = (c, d).
In Exercises 3740, copy and complete I he given statement so that it will
be a proportionality.
37. (5.7,10.12) = (10, [Sm, 0)
38. (0, H, 1) f (15* 37, 100)
39. (0,37. 0.67, 0.93) = ((?], 67, fTJ)
40. (V5 v^V^f CV&EH3)
396 Similarity Chapter 10
10.3 PROPERTIES OF PROPORTIONALITIES
As you might expect from your study of algebra and from some of
Exercises 10,2, proportionalities have some interesting properties. In
this section we show that the proportionality relation is reflexive, sym
metric, and transitive, and therefore it is an equivalence relation.
Hence the relation denoted by **=" has some properties in common
r
with the relations denoted by " = " and "^." We shall also state and
prove addition and multiplication properties. At the end of this sec
tion we consider some special proportionalites called proportions.
We shall prove the next two theorems for proportionalities involv
ing triples of numbers. It is easy to see how the statements and proofs
can be modified for proportionalities involving more than or fewer
than three numbers.
THEOREM 10.1 The proportionality relation is an equivalence
relation.
Proof:
1. The Reflexive Property. Let a, b* c be any real numbers. Ob
serve that (a, b f c) == (a, b, c) with proportionality constant 1.
2. The Symmetric Property. Suppose that {a, h % c) — (*£» e,/). Then
diere is a nonzero number k such that a = kd, b = ke t c = kf.
Why? Then d = k'a, e = k'b, f= k'c, where k' m l and
k' ffe 0. Therefore (d, <?,/) = (a, b, c). Thus if (a, b, c) = (d t ej)
with constant of proportionality k t then (d. e t f) = (a, b, c) with
constant of proportionality — .
3. The Transitive Property. Suppose (a, b* c) = (d t e t /) and
{d, e f f) = (g, h y i). Then there are nonzero numbers kj and k 2
such that
a = kid, b = k\e, c = k\f,
d = k 2 g, § = k 2 K f = M
Then
a = kid = fci(feg) as (hk 2 )&
b = k x e = ki(k 2 h) = (hk 2 )h,
c= kif=ki(k 2 i) = (k t k 2 )i.
10.3 Properties of Proportionalities 397
Since In jfi and k z # 0, it follows that Mfc f£ 0, Therefore, if
a, b t c are proportional to d, e, / with constant of proportional
ity &i, and d, e./are proportional to g, ft, i with constant of pro
portionality fca, then a, b, c arc proportional to g, h, i with con
stant of proportionality fci& 2 
In the next theorem note that (1) is an "addition" property and that
(2) is a "multiplication"' property.
THEORRV 10.2 If (a, b, c) =: (</, e f /), then
(1) (
and, if h ^ 0,
(1) (a, fe, c. a + b + c) = {d t e t f, d + e + f) y
(2) {ha t b>c) = {hd>e t f).
Proof: Let it be given that (a> b, c) = (ef t e, /). Then there is a non
zero number k such that
a = fcd, b = fee, c = Jfcf.
Adding, we get
a + b + c e faf + he + fcf,
and, by the Distributive Property,
a + b * c = k(d + e + /).
Multiplying both sides of a = kd by h, we get ha = h(M), and hence
ha =s Jt(fcd).
We have shown that if a. b % c are proportional to d, e, f with constant
of proportionality k, then (1) and (2) hold with the same constant of
proportionality.
Example 1 Figure 102 shows a triangle A ABC and a segment DE
joining a point D of AB to a point £ of AC. Suppose we know that
{AD, DB) = (A£, EC).
Figure 104
398 Similarity Chapter 10
We may conclude from the addition property of proportionality that
{AD, OB, AD + DB) = (AS, EC, AE + EC)
and hence that
(AD, DB, AB) = (AE, EC, AC),
since AB = AD + DB and AC = A£ f EC. Of course, we may also
conclude that
(AD, AB) j AE, AC).
This may not be the "whole truth/' but it is certainly the "truth." It is
like concluding that if x = 3», y — 3u, and z = Bm, then * = Bit and
y = Sv. Similarly, we may conclude that
(DB, AB) = (EC, AC),
Example 2 Let A ABC with D an interior point of BC be given as
suggested in Figure 103. Let /* denote the distance from A to BC, let
BD = bu and let DC = b* Then
 AABD\ = * 2 hbu \AADC\  $hb 2 .
Figure 104 B
Therefore (\AABD\, \AADC\) = (bi, bz), and the constant of propor
tionality is j/i. Thus the areas of the two triangles formed from A ABC
by inserting the segment AD are proportional to the lengths of the seg
ments formed from EC by inserting the point D, In this connection
some of you will recall Challenge Problem 29 in the Review Exercises
of Chapter 9,
We frequently work in geometry with proportionalities in which
two numbers are proportional to two numbers. Such proportionalities
are generally called proportions. Following a formal definition is a
list of some of their special properties.
Definition 10.2 If a t b, c, d are numbers such that
(a> b) =: (c, <l) is a proportionality, then that proportionality
is a proportion.
10.3 Properties of Proportionalities 399
lii other words, a proportion is a proportionality with two num
bers on each side of the " — " symbol.
The following theorem includes four named properties of
proportions.
THEOREM 10.3 Proportions involving nonzero numbers a, h,
c, d have the following properties:
1. Alternation Property. If (a, b) = (c, d), then
(a, c) = (b t d) and (cf, h) = (c, a).
2. Inversion Property. If (a , fr) =3 (e» d) % then
(h a) = (</, o).
3. Product Property, {a, b) = (c, d) if and only if ad = be.
4 Ratio Property, (a, b) = {c, d) if and only if y = ■£ .
p b a
Proof: Assigned as exercises.
If you think of a and d as the "outside" numbers and b and c as
the "inside" numbers of a proportion
(*. *) f te A
the Alternation Property says that if you interchange either the out
side numbers or the inside numbers in a proportion, then the result is a
proportion.
If you combine the Inversion Property with the Ratio Property,
the Inversion Property amounts to saying that if two ratios are equal,
then their "inversions" (reciprocals) are equal.
In Theorem 10.3, the numbers in the proportions arc required to
he nonzero numbers. What would lie the situation if zeros were per
mitted? Note that (0, 5) = {0, 10} is a true statement, whereas (0, 0) =
(5 T 10) is a false statement. Therefore the proportion
(0, 5) = (0, 10)
does not have the Alternation Property. Observe that « 7 s • 8 is a
true statement, whereas (0, 0) = (8, 7)isa false statement. Therefore
the Product Property does not apply to • 7 = ■ 8. Observe that
a,0) = (2,0)
is a true statement, whereas ^ = ^ is a false statement. Therefore the
Ratio Property does not apply to (1, 0) = (2, 0).
400 Similarity Chapter 10
EXERCISES 10.3
1. If 0, b t c t d are all nonzero numbers, then (a t h t c t d) — (a, &* c, d).
What is the constant of proportionality in this proportion? Which prop
erty of an equivalence relation does this illustrate?
2* Given {a, b, c, d) = faj\ g> h), prove that
fa f, g. h) = fa b> c d).
Which property of an equi valence relation does this illustrate? What
is the relation between the constants of proportionality for these two
proportionalities?
3. Given (a, b) = fa d) with proportionality constant k\ and (c, d) = faf)
with proportionality constant k>, then fa b) = fa f) with some pro
portionality constant, sny k^. How are k±, hit, kt related? Which prop
erty of an equivalence relation do we have illustrated here?
4. In the figure A, B, C are noncollinear points; A, B\, B& B are eollinear
and arranged in the order named; A, C u C 2i C are colUneur and ar
ranged in the order named. It is given that
(AB U »,B 3 . B 2 B) = (AC,, C,C 2 , dQ
with constant of proportionality 0.8. Prove, using the properties of
proportionalities, that
(a) (AB,,AB) = (AC lt AC).
(b) (16AB U AB) = (I6AC1, AC),
In Exercises 511, complete the given statement and name the property
which it illustrates. Assume that none of the numbers in these exercises is
zero.
5. If fa b) = (3, 5), then (a, 3) = (£>, 0).
6. lf(x,i/) = (6 3 7) t then ( ?/ ,,v) f (7,0).
7. If (m. c) = (5, 6) and (5, 6) = (x, tj), then [u, v) = (x, [?]).
8. If (a. b) = fa 00), then fa a + b) = fa c + d).
9. If (*, y) = (I, 2X then (3*. y) = {[[J, 2).
10. If (x, y) = (J7J, 7), then 7* = 4i/.
11. If (*, y) = (5, 8), then * = [B
10,4 Similarities between Polygons 401
In Exercises 1221, write a proportion, or complete the given one, so that
it will be equivalent to the given information. Starting with the given in
formation, you should be able to prove that the proportion is true. Starting
with the proportion, you should be able to prove that the given equation
(or equations) is true.
16. a = 3*. b = 3y 21. ^ = 
Exercises 2225 refer to Theorem 10,3. Prove that proportions involving
nonzero numbers have the indicated properties.
2SL The Alternation Property.
23. The Inversion Property.
24. The Product Property.
25. The Ratio Property.
10.4 SIMILARITIES BETWEEN POLYGONS
In this section we define what is meant by a similarity between two
polygons. In developing our formal geometry we use similarities as
tools; in most cases we consider similarities between triangles.
Definition 10.3
1. A onetoone correspondence ABC . , . * — > A'B'C , . .
between the vertices of polygon ABC . . . and polygon
A'B'C ... is a similarity between the polygons if and
only if corresponding angles are congruent and lengths of
corresponding sides are proportional.
2. If ABC . . . < — * A'B'C ... is a similarity, then polygon
ABC . . . and polygon A'B'C , , . arc similar polygons and
each is similar to the other,
3. If ABC . . . a — * A'B'C ... is a similarity with
A3 = kA'B', BC = kB'C, and so on, then k is the con
stant of proportionality, or the proportionality constant,
for diat similarity.
402 Similarity
Chapter 10
Notation, The symbol "W is read "is similar to"; hence ABC ,..*»
A'B'C ... is read "ABC . , . is similar to A'B'C . ..." This means that
A IH
A'B'C
Example 1 Fi give 104 shows two quadrilaterals ABCD and A 'B' CW
with segment lengths and angle measures as indicated. It ap
pears that ABCD * — * A'B'CD' is a similarity and hence that
ABCD ~ A'B'CD'. Let us check this conjecture. There arc eight pairs
of corresponding parts including four pairs of corresponding angles
and four pairs of corresponding sides:
LA
and I A'; Al5
and AT*';
B
and L W. BE
and JfC;
LC
and AC; UD
and CT?j
LD
and L D*; UA
and /XV.
D A
C
DO
1
I34X
90
1
X
Ag/^
A
7 B
P
FIgun KM
First we check the angles.
mZ.A = niLA' =
90,
mlB s mLB' =
45,
mLC = mLC =
135,
mLD = mLD' =
90.
Therefore, corresponding angles are congruent. Next we check to see
if the lengths of corresponding sides are proportional. We want
(AB, BC, CD. DA) = (A'B f . B'C, CD\ &A%
Substituting, we get
(7,3x/2*4, 3) = (3.5, 1.5 y/2, 2, 1.5).
Since this is indeed a proportionality, with constant of proportion
ality 2, the lengths of corresponding sides arc proportional. There
fore, it follows dircctJy from the definition of a similarity that
ABCD A' B'C D\
10.4 Similarities between Polygons 403
In Example 1 the constant of proportionality between the lengths
of the sides of quadrilateral ABCD and the lengths of the correspond
ing sides of quadrilateral A'B'CD' is 2. Is it possible that in another
example the constant of proportionality might be 1? Of course it is. In
this special case of a similarity, corresponding angles are congruent
and corresponding sides are congruent. Hence for this special case you
should sec that Lhe similarity is a congruence. In other words, a
congruence between polygons is a similarity between polygons for
which lhe constant of proportionality is L
Just as congruence for triangles is an equivalence relation , so also
is similarity for polygons. We slate this fact as our next theorem,
THEOREM 10 A The relation of similarity between polygons is
reflexive, symmetric, and transitive.
Proof: We shall prove the theorem for triangles. Tt is easy to modify
this proof to get a proof for quadrilaterals* pentagons, and so on.
Reflexive. Let AABC be given. Since AABC at AABC, it follows
that A ABC — A ABC. What is the constant of proportionality for
(AB, AC, BQ = (AB, AC, BQ?
Symmetric. Suppose that AABC — ADEF; then ZA^ZA
ZB^ZJE, ZC==ZF,and
{AB, BC, CA) = (DE, EF, Dl<)>
We want lo prove that ADEF — AABC, which means that ZD^
LA, LE^i LB, LF= / C, and that
(DE, EF, DF) = (AB, BC, CA).
Now LA=*LT) implies LD^LA, LBsLE implies LM m£B,
and £Cm£W implies LF = Z C (Which property of congruence
for angles supports tills deduction?) Also,
(AB, BC, CA) = (DE, EF, DF)
implies Hi at
fVm, EF, DF) = (AB, BC, CA).
(Which property of proportionality supports this deduction?)
Therefore
AABC — ADEF
implies that
AD£F AABC,
and similarity for triangles is a symmetric relation.
404 Similarity Chapter 10
Transitive* Let it be given that AABC ~* ADEF and that
ADEF — &GHL The proof that A ABC ~ AGH1 is assigned as an
exercise. From tills it follows that similarity for triangles is a transitive
relation,
THEOREM 10,5 The perimeters of two similar polygons arc
proportional to the lengths of any two corresponding sides.
Proof: As in the case for Theorem 10.4, we shall prove Theorem 10.5
for triangles. It is easy to modify this proof to get a proof for quadri
laterals, pentagons, and so on,
Ijet A ABC — AA'B'C be given and let p be the perimeter of
A ABC and p* the perimeter of AA'B'C, By the definition of similarity
we have
(AB, BC, CA) = (A'B\ B'C, CA').
By Theorem 10,2,
(AB, BC,AC, AB + BC+ CA)
= {A'B\ B'C, A'C, A'B' + B'C + CA').
Bui
p = AB + BC + CA and p* = A'B' + B'C + CA',
Therefore
(AB, BC, CA. p) = (A'B', B'C, CA\ p%
It follows that
(AB, p) = (AW, jf)
and, from the Alternation and Inversion Properties that
(p,p') = (AB.A'ff).
In a similar way, it can be shown that
(p, p') = (BC, B'C) and (p, p') f (CA, CA').
This completes the proof of Theorem 10.5 for triangles.
EXERCISES 10.4
1. Complete Ihe proof of Theorem 10.4 by proving thai
A ABC AGHl
2. Given A ABC  ADEF, mLA = 30, m£B = 60, AB m 20,
BC = 10, CA = 10 V3, DE = 100, find mlD, mlE, mZF, EF,
and DF,
10,4 Similarities between Polygons 405
In Exercises 310, ihcre are figures showing two similar triangles with some
side lengths indicated and a similarity statement given. In each case, find the
proportionality constant of the given similarity statement and Uks lengths
of any sides whose lengths are not given.
3.
A ABC 6D£P
4, A
a P 7 S S q
A ADC  AACB
&PQR ATQS
7, T
ATSZ  AXV7
406 Similarity
8. D
Chapter 10
9. J?
t l A
ARKD A HAT
&ABC  AA'R'C
11. Given: A ABC <~ AA'B'C
AA'B'C — AA"£"C"
AB = JA'S*
A'W  5A"B"
Pmce: AABC =g AA"J?"C"
12. Let AAXT — AAtt/D wilh
A\ AT AT
HU AfD DD
= 1000.
If the shortest side of AMTflD is of length 1000, find the length of the
shortest side of A A AT.
10,4 Similarities between Polygons 407
13. II in Exercise IS, p and p' denote the pcriinclcrs of the smaller triangle
and tile larger triangle, respectively, find *— ■
P
14. ABCDE and A'B'UD'E' are pentagons such that
(AB t BC, CD, DE, EA) = (A'B\ WC t CD f t &E% E'A%
£A=*£A\ LBzt£W t £Cm£O a
LD^LD\ £Bat£&, and AB = ISA'S 1 .
Prove that the perimeter of ABCDE h 13 times the perimeter of
A'B'CD'E'.
15. If A ABC ~ AA'B'C and AB = 10, HC = 8, A'fl' = 25, A'C = 35,
find B'C and AC.
16. If AFQR ~ ASTV and PQ = 24, ST = 16, PR = 18, TV = 10. find
Qfl and SV.
17. If A ABC ~ ARST and 7 ■ AB = 4 « RS, what is the ratio of the per
imeter of A ABC to the perimeter of ARST?
18. If A ABC ADEF, AB = 5, BC = 7, AC = 8, and the perimeter
of ADEF is 60, find DE, £F t and DF.
19. In the figure, AKTS, RK ± S3, RK = 3. AT = 8, 15 = 4.
(a) Find  AHAT, the area of A RAT.
(b) Find AflTS,
(c) Solve for at: ( A RATI, ARIS) = (,v, 4)
20. In the figure, AKTS> M. 1 A3, JW: = 3, AT = 12, TS = 2.
(a) Find [ARAT.
(b) Find ,AHTS.
(©) Solve for *: (*, ARTS) = (12, 2).
408 Similarity Chapter 10
21. In the figure. AfSOT, M J. S7T, MO = 12, OT = 5.
(a) Solve for xs flAiVAfq, A.VOTJ) = (12, *).
(b) Solve for tjx (\ANMO\. y) = (A.VOT P 5).
In the figure, RFST, RDEW, SE  TW t FE ± KT, D5 ± EW,
RS = 20, ST = 10, RE = 30, EW = 15, FE = 29.
(a) Find \ARSE\, (b) Find  AST£.
(c) Find SD, (d) Find  A SEW].
In the Bgure, ADB, AEC, Z5E  BC, £D = 12, DA
\&BDE\ = 120.
(a) Find;AADE.
(b) FirtdjADECj.
(c) Find the ratio of AE to EC.
(d) Find the ratio of AD to DB.
(e) Compare the ratios in (c) and (d).
10,5 Some Length Proportionalities 409
10.5 SOME LENGTH PROPORTIONALITIES
In this section there arc several theorems regarding the lengths of
segments formed by lines parallel to one side of a triangle intersecting
the other two sides, and there are other theorems extending these ideas
to lengths of segments formed when three or more parallel lines are
cut by two transversals. These theorems are useful in proving the
similarity theorems of Section I0;6.
THEOREM 10.6 (Triangle Proportionality Theorem) If a line
parallel to one side of a triangle intersects a second side in an in
terior point dien it intersects the third side in an interior point, and
the lengths of the segments formed on the second side are propor
tional to the lengths of the segments formed on the third side.
Proof: Let there be given a triangle A ABC and a line / such that I
intersects KB in an interior point D and such that t is parallel to EC as
suggested in Figure 105. We know from Theorem 2.6 that Winter
sects A(7 in an interior point Call it £. We shall prove that
(J5D, AD, BA) = (CE, A£, CA).
A
Figure 105
Ut
hi be the distance from E to AB,
/*2 be the distance from D to AC t
ft 3 be die distance between t>E and BC.
Following are the main steps in the deductive reasoning which com
pletes the proof of the theorem. You are asked in the Exercises to sup
ply the reasons for steps 1 through 7 (definitions, theorems, preceding
steps; etc.).
1.
BD ^hjBD _ ABED\
AD ~ ^i t  AD " \AAED\
_ CE MioCE _ \ACED\
* AE " %hg'AM ' AAED
3.  ABED\ = ^ 3 • DE = \ACED\
410 Simitar ity Chapter 10
4 BD = CE
' AD AE
5. BDAE = AD'CE
6. (BD, AD) = (CE t AE )
7. {BD, AD, BA) = (CE, AE, CA)
Therefore, the lengths of KB and the segments formed by I cutting
KB are proportional to the lengths of 7l€ and the segments formed
by I cutting JlC. From step 7 we get step 8,
8. {BD, BA) s (CE, CA) and (AD, BA) = (AE, CA)
From step 8 we got step 9 using the Alternation Property.
9. (BD, CE ) = (BA, CA) and (AD, AE) = [AB, AC)
Therefore, the lengths of AB and the segments funned by I cutting
other two sides in interior points, then it exits oif segments whose
lengths are proportional to the lengths of those sides.
THEOREM 10.7 (Convene of the Triangle Proportionality
Theorem) Let A ABC with points O and E such that ADB and
AEC be given. If
(AD t AB)~(AE t AC),
then DE I BC.
Proof: Let AABC with points D and £ such that ADB, AEC,
(AD t AB) =; (AE t AC), as suggested in Figure 1Q6, be given.
B
Figure MMSI
Let I be the unique line through D and parallel to BC. Let W be the
point in which ] intersects AT. In the figure, E and A" appear to be dif
ferent points. We shall show that they are actually the same point.
Since I is parallel to BC, It follows from Theorem 108 that
(AD,AB) = (AE\AC).
10,5 Some Length Proportionalities 411
But it is given that
Therefore
Then
(ADtAB) == (AE.AC).
(AE\AC) = (A£, AC).
Why?
AE'AC = AEAC (Why?)
and AE' = A£. Since E and £' arc points of AC that are on the same
skle of A and at the same distance from A* it follows that £ = E'. Since
/ = ££' = M t
it follows that BE  5C.
The word "cut" is used frequently as a synonym for "intersect/'
particularly in situations involving several lines and a transversal,, as in
our next theorem.
THEOREM 10.8 If two distinct transversals cut three or more
distinct lines that are coplanar and parallel, then the lengths of the
segments formed on one transversal are proportional to the lengths
of the segments formed on the other transversal.
Proof: We shall prove the theorem for three p. ir.il lei lines. It is easy
to modify the proof for more than three parallel lines.
V
Figure 107
Let * and / be two distinct transversals of three distinct lines (, m,
n that are coplanar and parallel as suggested in Figure 107. Label the
points of intersection as in the figure. Draw XF. Let G be the point
of intersection of m with AF. It follows from the Triangle Proportion
:tlt(\ Tli or .rem. applied to ;\ACh, rhat
I:
{AB > AC) = (AG t AF)
and from the same theorem, applied to &AFD, that
(2)
(GA, FA) = (ED, FD).
412 Similarity Chapter 10
From the Transitive Property of Proportionality it follows that
:V3) (AB t AC) = (DE,DF).
Similarly, we may show that
(4) (BC t AC) = (EF,DF).
Then from (3) and (4) we gel
(5) (AB, BC, AC ) = (DB, EF t DF ).
COROLLARY 1Q.8J If a line bisects one side of a triangle and is
parallel to a second side, then it bisects the third side.
Proof: Assigned as an exercise*
EXERCISES 103
In Exercises 15, there is a triangle, &ABC t with points D and £ such that
ADB and AEC, 75E  BC, and with lengths denoted as follows:
p = AD, q = DB, rmAE,tm EC In each case, given three of the four
numbers p. q, r, s, find the missing one. Draw and label a figure for each of
these exercises,
1. p = 5, q a 6, r = 75, * = [7]
2. p = 8, q = G> r = 0, « = 9
3. p = 18, <f = (TJ, r = 10, * = 13}
4. p = [0, </ = »$, t = 7, * = 20
5. p = V£ </ = "\A *" = 4 * =
In Exercises 610, there is a triangle, A ABC, with points and E such tliat
AIKBaudAEC, and with lengths denoted as follows: x = AH,jj = AC T
p — AD, q = DB, r = AE, tt = EC. In each case, given some of these
lengths, determine whether or not the lines DE and BC are parallel. Draw
a figure for each exercise.
6. x = 10,
if = is.
p = 5,
r=S
7. * = 10,
!/ = 15,
p = 5,
r=7.5
8. p = 25,
q = i5,
r = 60,
* = 3fl
9. p a 0.9,
s = 0,88,
r = 0,81.
q = 0.81
10. p = ^
* = 2,
r=2,
t/ = 4
11. Write a "reason" for each of steps 1 through 7 in the proof of Theorem
10.8,
10.5 Some Length ProportDomlitit* 413
12. Coplan.tr and parallel lines a, b, c r d arc cut by transversals s and f sis
•ii i'."/' ; Mn the figure, Given Lengths C& > cements SS labeled in Lite lim
ine, find x and tj.
_^
■fro
M
5.4
¥d
13. If tlirer distinct c:oplanar and parallel lines arc cut by two distinct par
allel transversals, then the lengths of the segments formed on one trans
versal are proportional to the lengths of the segments formed on the
other transversal. What is the constant of proportionality in this case?
14. In the proof of Theorem 10.8, Figure 107 suggests that s and / do not
intersect in the portion of the plane between lines / and n. Draw a figure
for this theorem which shows ,v and * intersecting at a point between
lines / and n. b the proof for this theorem, as given, applicable for the
case suggested by your figure?
15. Consider again Figure 107. Suppose that the figure is modified to show
s and f intersecting at a point P on the opposite side of hne / from it
Using Theorem 10.6 and Figure 107 suitably modified and labeled*
obtain some proportionalities involving lengths of segments with P as
one endpoint, Use these proportionalities to prove that
(AB, AQ = (DE, DP).
16. Complete the following proof of Corollary' 10.8.1. Let A ABC be given
with D the midpoint of AB and let I he the line through D, parallel to
BC, and Intersecting A Cat E as shown in die figure. Let l\ be the line
through A and parallel to J. Then l x  BC. Why? Then
{AD t DB) = (AE, EC). Why?
It follows that AD_ _ AE _ , .„. „
DB ~ EC~ '
Complete the proof by showing that / bisects AG.
A
4 —
d/
\e
Hi
1
414
Similarity
Chapter 10
17.
(An in
of con
length
tractoi
Copy
formal geometry exercise.) With the aid of a rale
i passes, draw two triangles, A ABC and AA'B'C
of the sides (in centimeters) are as listed in the table.
; measure the angles A, B, C, A', W, C to the n©
the table and complete it by recording the angle in
r and a pair
so that the
Using a pro
test degree.
fiisurev
AABC
AA'B'C
AB = 9.0
A'W = 13.5
BC = 7.0
ffC = 10.5
AC = 6,0
A'C = 9.0
m£A = [£J
mlB = [?J
roZC = [T
m/.A' = g]
mLW m E3
mlC = 1]
1& {An informal geometry exercise.) With the aid of a ruler and protrac
tor, draw two triangles, A ABC and AA'B'C, with side lengths (in
centimeters) and angle measures as indicated in the table. Measure the
remaining parts of the two triangles and record the results. Measure
lengths to the nearest 0,1 cm. aud angles to the nearest degree.
AABC AA'B'C
AB = 8.0 A'B' = 10.0
BC = 10.0 JTC = 12.5
AC = CO A'C = [7j
mlA = {jl m£A' = {2]
mZB = 46 m/.B' = i6
i«ZC = (T) inZC = [TJ
lfc {An informal geometry exercise.) With the aid of a ruler and a pro
tractor, draw two triangles* A ABC and AA'B'C with side lengths (in
centimeters} and angle measures as indicated in the table. Measure the
remaining parts of the two triangles and record the results. Measure
lengths to the nearest 0.1 cm. and angles to the nearest degree.
AABC AA'B'C'
AB = 10 A'B' = 6
BC = CD B'C = \T]
AC=\?] AC =
mZA = 32 mZA' = 32
mZB = 51 mLW = 51
mlC=\2} mZC = [?3
10. 6 Triangle Similarity Theorems 41 5
10.6 TRIANGLE SIMILARITY THEOREMS
In our study of congruence for triangles we first defined congru
ence so that, by definition, all six parts of one triangle must be congru
ent to the corresponding parts of a second triangle in order for the
triangles to lie congruent to each other. On the basis of our experi
ences with triangles it seemed reasonable to expect all six of the re
quired congruences involving sides and angles to be satisfied if certain
sets of three of them are satisfied So we adopted the wellknown Tri
angle Congruence Postulates, referred to as S.A.S., A,S,A., and S.S.S,
Similarly, our experiences with triangles, especially the triangles of
Exercises 17, 18, 19 of Exercises 10,5, suggest that, if certain combina
tions of some of the definitional requirements for a triangle similarity
are verified, then all of the requirements for a similarity are satisfied.
Since we adopted Congruence Postulates, it would seem reasonable to
adopt Similarity Postulates. It turns out, however, that it is not difficult
to prove what we want to know about similarity; hence in this instance
postulates are not necessary. First, we prove a theorem that is useful in
proving the main Similarity Theorems.
THEOREM 10.9 If A ABC is any triangle and k is any positive
number, then there is a triangle AA'B'C such that AA'B'C *—
A ABC with constant of proportionality k.
Proof: Let triangle A ABC and a positive number k be given. We
consider three cases.
Case 1. k <C \.
Case 2. k = L
Case 3. k> 1.
We shall prove the theorem for Case 1 and assign the other two
Cases in the Exercises. a
Suppose that k < 1; then there Av
is a point D on AB such that / ^s^
AD = k* AB and a point E on p/— — .J>.
£C such that AE = k*AC In / / ^\
Figure 108, DE is drawn so that g f ^5,
k appears to be about 0,6 , Figure io8
It follows from the converse of the Triangle Proportionality
Theorem that DE , EC, It follows from theorems regarding parallel
lines that
LADE & L B and LA2ED ss L C.
416 Staiilirfty Chapter 10
Of course. LA^ I A. As you might expect, it is AADE that qualifies
as it suitable AA'FC, that is, ADE «— * ABC is a similarity. So far
we have shown that corresponding angles are congruent and, from the
way we have chosen D and E t we know that
(AD, AE) = (AB y AC).
We need to show that
{AD, AE, DE) = (AB, AC. BC).
_Let F be the point of BC in which the line through £ and parallel
toJB intersects BC. Then DE = BF(Why?) and
{AE, BF) y (AC t BQ. Why?
Substituting, we get
(AE, DE) = (AC\ BQ.
From tins proportion and the preceding proportion,
(AD,AE)j(AB s Aq,
it follows that
(AD, DE, AE) = (AB t BC, AC),
which completes the proof for Case 1 in which k < 1.
THEOREM 10.10 (S,S.S. Similarity Theorem) Given A ABC
and ADEF, if
(AB, BC, CA) = (DE, EF, FD),
then A ABC— ADEF.
Proof; Let A A BC and A DEF such that
(AB, BC, CA) = [DE, EF, FD)
he given. (Sec Figure 109.) Suppose the constant of proportionality
is k. From Theorem 10.9 it follows that there is a triangle AD'E'F such
that AD'E'F *— ADEF Witt) proportionality constant ft. Then
(AB. BC, CA) = (DE. EF, FD) with proportionality con
stant k.
(&E\ FF, FD) = (DE, EF, FD) with proportionality con
stant k,
(AB,BC,CA) = (D , E\ET,F'D f ) with proportionality con
stant 1. Whv?
10.6 Triangle Similarity Theorems 417
D* *F D'
Figure 109
From this we conclude that AB = DT', BC = E'F, and CA =
Ft/. It follows from the S.S.S. Congruence Postulate that
AABC== AD'E'F,
BecaU now that triangle congruence is u special case of triangle simi
larity and that triangle similarity is an equivalence relation. There
fore AABC AD'E'F. But AD'ITADEF, it follows thai
A ABC— AD£F,
THEOREM 10.11 (S.A.S. Similarity Theorem) Given AABC
and ADEF, if
Z A =s LD and (AB, AQ = (DE, DF),
then AABC ADEF.
Proof: Let AABC and ADEF be given with L A ss Z D and
(AB, AQ = (D£, DF).
(Use Figure 109 again.) Suppose die constant of proportionality is k.
Let AD'E'F be a triangle such that AD'F/F ~ ADEF with propor
tionality constant k. Then
AB = fcDE, D'£' = Jfc«D£ t AB = WB %
and
AC? = k ■ DF, DT = Jt ■ DF t AC = D'F.
It follows from the S.A.S. Congruence Postulate that
AABC = AD'E'F.
Then AABC ~ AD'E'F and AUWF ~ ADEF, and wc may con
clude that AABC  ADEF.
COROLLARY 10.11.1 A segment which joins the midpoints of
two sides of a triangle is parallel to the third side and its length is
half the length of the third side.
Proof: Assigned as an exercise.
and
AA=*AD f ;
and
IB^AF/
and
AB = D'E'.
418 Similarity Chapter 10
THEOREM 10.12 (A.A. Similarity Theorem) Given A ABC and
ADEF, if A A s Z D and ZB ss Z E, then AABC — AJ9£F.
Proofi Let AABC and ADKFsuch liiat LA m AD and AB~z AE
he given. (Use Figure 109 once more.) Let
AB _ k
Let AD'E'F be a Iriangle such thai AD'E'F * ADEF with propor
tionality constant rf, Then
AA^ID, ADsiAD\
ABmAM, AEatAE\
AS = kDE, DT = kDE,
It follows from the A.S.A. Congruence Postulate that
AABCs ADTF.
Then A ABC  AD'E'F and A£>E'F  ADEF, and we conclude
that A ABC — ADEF.
Note that we have an S.S.S. Congnience Postulate and an S.SS.
Similarity Theorem f and that we have an S.A.S. Congruence Postulate
and an S.A.S. Similarity Theorem, hut that we do not have an A.S.A.
Similarity "Theorem to match our A.S.A. Congruence Postulate. Of
course, we could, if wc wished, call our A.A. Similarity Theorem die
A.S.A. Similarity Theorem. But if Z A a* AD and AB =* AE, we do
not need to be concerned, about whether "AB is proportional to DE,"
Indeed, if AB and DE are any two positive numbers whatsoever, there
is a number k such that
ABa It JOB.
Look at the tables you prepared for Exercises 17, 18, 19 of Exercises
10.5. Do the measurement data recorded in the tables illustrate the
triangle Similarity Theorems? They should. Which theorem does Excr
eta 17 illustrate? Which theorem docs Exercise 18 illustrate? Which
theorem does Exercise UJ illustrate?
We have written the triangle Similarity Theorems using quite a few
symbols. Is it possible to state them in a more relaxed form without
symbols? In the following versions of the theorems we use the word
^corresponding" without "pinning it down." It should be understood
in each case that a correspondence between the vertices of one triangle
and the vertices of the other triangle is fixed so that there are corre
sponding parts.
10.6 Triangle Similarity Theorems 419
THEOREM 10.10 (S.S.S. Similarity Theorem— Alternate Form)
If the lengths of the sides of one triangle are proportional to the
lengths of the corresponding site of the other triangle, then the
triangles are similar.
THEOREM 10.11 (S.A.S. Similarity Theorem— Alternate Form)
Tf an angle of one triangle is congruent to an angle of another tri
angle and if the lengths of die including sides are proportional to
the lengths of the corrcspondint; sit Irs in the oilier triangle, then
the triangles are similar.
THEOREM 10J2 {A. A. Similarity Theorem— Alternate Form)
If two angles of one triangle arc congruent to the corresponding
angles of another triangle, then the triangles are similar.
The following theorem points out that if the sides of one triangle
are parallel to the sides of a second triangle, then the two triangles are
similar. The property of parallel sides is a sufficient condition to ensure
similarity* but, of course, it is not a necessary condition.
THEOREM10.13 U triangles AABC and ADEF are such that
A~B .1 m. BC I! EF, UA II FD, then AABC  ADEF.
Proof: Assigned as an exercise.
EXERCISES 10.6
1. Prove Theorem 10,9 for the case in which k = L
2. Prove Theorem 10.9 for the case in which k > 1.
In Exercises 3 and 4. two triangles and the lengths of their sides are given
by means of a labeled figure.
a b AABC — APQR?
Is AABC ~ AQPR?
h AABC APRQ?
Is AABC— ARPQ?
Is AACB ~ ARQF?
Is ACBA  AQPR?
4. Is AABC— A CDE?
h AABC— ADEC?
Is AABC AEDC?
Is A CAB A DEC?
Is AC/iA — ADCEr
Is ABAC— A DEC?
420 Similarity
Chapter 10
In Kxerciscs 5 and 6, two triangles arc given in a figure wilh some segment
lengths and angle measures.
5.
6.
Is AABC
Is AABC
Is AABC
Is AABC
Is Aj\BC
Is AABC
ADEF?
A EFD?
ADFE?
AEDF?
AFED?
AFDE?
Is AABC ~ ADEB?
Is AABC— ADKE?
Is A A KG— ABED?
Is AABC— &JHBEP
Is AABC— AEBD?
Is AABC ~ AEDB?
7. Given isosceles AABC with AB ~ AC* and with points D, /£, Fsueh
dnit ADB, JJEC, CFA, DE±AB, FE 1 XC, prove that
ABD£— ^C3^.
8, If at a certain Lime, in a certain place, a certain tree casts a shadow 40 ft,
long and a 6ft, man casts a shadow 2 ft. and 3 in. long, find the height
of the tree.
Exercises 911 refer to the figure with AEC, BDC, ED  A~B, and seg
ment lengths as marked.
9. Name a pair of similar
triangles and explain
why they are similar.
10, Find r.
11, Find*
12, In the figure, ADC and
LABC^lBDC. Name a
pair of similar triangles and
explain why they are similar.
10.6 Triangte Similarity Theorems 421
D C
13. In the figure* AB  CD,
IDE X AC, W±7kC t AEF,
and EFC. Name a pair of
similar triangles and explain why
they are similar.
Exercises 1420 refer to the figure with BTJ , OE, DB  EC, ABD, DEF,
ACF f and A~F X F25.
14. Prove AAFD ~~ &CFE.
15. Prove ACF£ — AACB,
16. Prove AAFD — AACB,
17. If CF = 2, BD = 3, and
AC = 8, find AB.
18. If AB = 12, BD = 3, and
BC = 8, find DF
19* If AD = 18, AF = 9, and AC = 7, find EC.
20. If CF = 3, KF = 4, and AC = 7, find DF.
21. In the figure, ^fE and CD are altitudes
of AABC, AFE, and CFD.
(a) Prove AB£A  A.BDC.
(b) Prove AAW ACEF.
22. Given AAJ3C with points D and
£ such that ADB, BEQ
AE X BC, CD X SB, prove that
(AEtEfyiCDrDB).
23. Given two right triangles, A ABC and
AAPQ, as in die figure, copy and com
plete the following proportionality involv
ing lengths of the sides of these triangles:
{AB.BC,CA) = (AP T [2],[I]).
24. Given AABC  ADEF, AB = §, BC = 7, AC = 10. DE = 7, fi
EF and DF
422 Similarity
Chapter 10
25. Given A ABC  APQH prove dial if A ABC is a right triangle, then
APQR is also a right triangle,
26. Prove Corollary 10.11.1.
27. CHALLENGE PROBLEM, CiVCIl
parallelogram ABCD with
BEC, AE and ED intersect
ing at h\ and BF = $BD,
prove that BE = £  BC
28. Prove Theorem 10.13. Consider two cases: (a) AB and DE are parallel,
BC and EF are parallel, CA and VD are parallel, and j b) AM and 15 JS are
antiparailcl, BC and /Li*' are antiparaUel, CA and F£) are an ti parallel Use
Theorems 7.26 and 7.28.
10.7 SIMILARITIES IN RIGHT TRIANGLES
Sometimes base and altitude are interpreted as segments and some
times as numbers (lengths of segments). In our next theorem altitudes
are segments If AABC ~ AA'B'C, then we have agreed that A and
A' are corresponding vertices, KB and A'B' are corresponding sides,
and so on. It is natural to extend this idea to include corresponding
altitudes, that is, altitudes from corresponding vertices.
THEOREM 10.14 If two triangles are similar, then the lengths of
any two corresponding altitudes are proportional to the lengths of
any two corresponding sides,
Proof: Given A ABC ~ AA'B'U, let D and D' be the feet of the
altitudes from A to E€ and from A' to B'C, respectively. Let a = BC,
b = CA, c = AB, h = AD, a' = B'C, b' = CA\ c' = A'B',
h' = A'D'.
We shall prove the theorem for the case in which BDC, as shown
in Figure 1010, The remainder of the proof is assigned in the Exercises .
>1^
Figure 1010
10,7 Similarities In Right Triangles 423
Since BDC, Z B and Z C are acute angles. Why? Then Z B' and
Z C are acute angles. Why? Then it is impossible for D' to be either
the point £' or the point C\ Why? Also, it is impossible that D'B'C
or that WCD'. If ETWC, then AB'&A' is a right triangle with an
acute exterior angle contrary to the Exterior Angle Theorem. There
fore B'DC as indicated in Figure 10.10,
Now /_B ^ LB {Why?) and LBDA ^ LB'D'A* (Why?). It fol
lows from the A.A. Similarity Theorem that AADB — AA'D'B'.
Therefore
But
(c. ft) = (c', A').
(a, 6, o) = (a', b f , c'l
Therefore
(a, c) == (a\ c') and (fo, cj = (//,
It follows from the Alternation and Inversion Properties of Propor
tions that
{K /i') f (c, v'% (c, c') = (a, a% and (c, c<) f (ft, I,*).
It follows from the Equivalence Properties of Proportionalities that h
and h' are proportional to the lengths of any two corresponding sides.
THEOREM 10,15 If two triangles are similar, then their areas
are proportional to the squares of the lengths of any two corre
sponding sides.
Proof; Given A ABC ~ AA'iTC, let D and D' be the feet of the
altitudes from A to W and from A' to B'C, respectively. Let a = BC,
h = CA, c = AB r h = AD, a' = B'C. 1/ = CA\ tf = A'B\ h r =
A'D'. (See Figure 1011.) It follows from Theorem 10,14 that
A
B D
B ir
Figure 1011
424 Similarity Chapter 10
Suppose that h = ka; then h' = kif and
\AABC\ = {ah = %a(ka) = $k)a* t
 AA'B'C' 1 = \u'h' = \<f{W) = (JfttfA
Therefore the areas of A ABC and AA'jB'C are proportional to a unci
(rt') 3 . Similarly, it may be shown that the areas are proportional to h 2
and (by and'to c 2 and (c') a .
THEOREM 10.16 In any right triangle the altitude to the hypote
nuse separates the right triangle into two triangles each similar to
the original triangle, and hence also to each other.
Proof: Let A ABC be a right triangle with the right angle at C and
with D the foot of the altitude to the hypotenuse. Then D ,/ A and
D / B. (Which theorem is the basis for this assertion?) Also* it is iin
possible to have DAB or ABD. (If either of these betweenness re
lations ig true, there is a triangle with D as one vertex with one interior
angle a right angle and one exterior angle an acute angle. Which the
orem does this contradict?) Therefore D is an interior point of AB as
suggested in Figure 1012.
Figure 10 IS
You are asked in the Exercises at the end of this section to complete the
proof by showing that A ABC, AACD, and I\CBD arc all similar to
each other.
Next we have two corollaries that follow easily from Theorem
KXIG, but first we need some definitions.
Definition 10.4 If P is a point and I is a line, the projection
of P on Ms (1) the point P if F is on / and (2) the foot of the
perpendicular from P to I if P is not on /.
Definition 10.5 The projection of a set S on a line / is the
set of all points Q on t such that each Q is the projection on J
of some P in S.
10.7 Similarities in Right Triangles 425
Compare Definitions 10.4 and 10.5 witb Definitions 8,9 and 8.10
in which we denned a projection on a plane.
Note in Figure 1012 that AD is the projection of AC on AB. In
deed, A is the projection of A T D is the projection of C, and every point
Q such that AQD is the projection of a point P such that APC,
Conversely, every point P such that APC has as its projection on AB
a point Q such that AQD.
Since the projection of AT on AB is a part of the hypotenuse in the
situation of Theorem 10.16, we may say that AD is the projection of
AC on the hypotenuse. Similarly, 7M is the projection of CB on the
hypotenuse.
COR OLLAR Y 10J0.1 The square of the length of the altitude to
the hypotenuse of a rigfat triangle is equal to the product of the
lengths of the projections of the legs on the hypotenuse.
Proof; In Figure 1012, AD is the projection of AC on AB, and UB
is the projection of WC on AB. In the notation of the figure, we must
prove that
(CUP^ADDB.
Using Theorem 10.16 and some properties of proportionalities, wc
have
AACD  ACBD
(AC, CD, AD) = (CB f BD, CD)
{CD, AD) = (BD, CD)
(CD)* = AD*DB
COR OLL\R Y I 0. 1 &£ The square of the length of a I eg of a right
triangle is equal to the product of the lengths of the hypotenuse and
the projection of that leg on the hypotenuse.
Proof Assigned as an exercise.
Definition 10.6 If a and b are positive numbers such that
(a, x) = (*, b)
or that
fe *) f ft *).
then % is called a geometric mean of a and h.
426 Similarity
Note that if
or if
Chapter 10
(*. x) f fe b)
(x, a) = (h, 4
then x 2 = ab (Why?) and ar = y/aB or x = — yS5. We often call y/aB
the geometric mean of a and b. In view of Definition 10,6, Corollary
10.16 .1 and Corollary 10.16.2 can be restated as follows,
COROLLARY 10.16.1 {Alternate Form) The length of the alti
tude to the hypotenuse of a right triangle is the geometric mean of
the lengths of the projections of the legs on the hypotenuse.
COROLLARY 10A6.2 (Alternate Form) The length of a leg of
a right triangle is the geometric mean of the lengths of the hypote
nuse and the projection of that leg on the hypotenuse.
In Chapter 9 we proved the Pythagorean Theorem using proper
ties of areas and suggested two other area proofs in the Exercises. One
of the shortest proofs of the Pythagorean Theorem is an algebraic proof
that follows easily from Corollary 10.16.2. We state the Pythagorean
Theorem again and outline a proof that employs Corollary 10, 16.2, We
also proved the Converse of the Pythagorean Theorem in Chapter 9.
We state the converse again; we shall not prove it again.
THEOREM 10,17 (TJte Pythagorean Theorem) In any right tri
angle the square of the length of the hypotenuse is equal to the sum
of the squares of the lengths of the two legs.
Proof: Let A ABC With a right angle at C be given. (See Figure
1013.) Let D he the foot of the altitude to the hypotenuse AT?. Ixt
AR = c,BC = a, CA = b. AD a x t twdDB == c  *. Then it follows
from Corollary 10.162, with a the length of a leg and c — x the length
of its projection on the hypotenuse that a 2 = (c — x)c t and with b the
length of a leg and x the length of its projection on the hypotenuse that
b 2 =. xc .The proof may be completed by .showing thala? ■+ fr 2 = c 2 .
Figure 1013
A
* D
10.7 Similarity* in Right Triangles 427
THEOREM 10.18 {Converse of the Pythagorean Theorem) If
a 2 + fc 2 = c 2 , where a, h, c are the lengths of the sides of a triangle,
then the triangle is a right triangle with c the length of I he
hypotenuse.
EXKKCJSES 10.7
I, Given right triangle A ABC with hypotenuse AB, let D he the foot of
the altitude to EC. E the foot of the altitude to AC, and P the foot of the
altitude to AB, How many distinct points are there in the set
(A,B, C,«, E.F)?
In Exercises 27, there is a figure show mga right triangle with hypotenuse
AT3 and with D the foot of the altitude to aB, In each case, given the lengths
of some of the six segments, find the lengths of the other segments. Express
your answers in exact form using radicals if necessary.
3. B
C 4
4, b
428 Similarity Chapter 10
H, Find the perimeter of an equilateral triangle if the length of each of its
altitudes is 10.
9. Find the length of the diagonal of a rectangular floor to the nearest foot
if the floor is 21 ft. wide and 28 ft. long.
10. A ladder 12 ft long reaches to a window sill on the side of a house. If
the window sill is 9 ft. at»ove the (level} ground, how far is the foot of
the ladder from the side of the house?
11. Find the length of the hypotenuse of a right triangle if its legs each have
length 1. Express the answer exactly using a radical.
12. Find the length of the hypotenuse of a right triangle if its legs each have
length 100.
13. Find the lengths of the legs of an isosceles right triangle whose hypote
nuse has length 1.
14. Find the lengths of the legs of an isosceles right triangle whose hypote
nuse has length 2.
15. Find the lcugtlis of the legs of an isosceles right triangle if the length of
its hypotenuse is \/2.
16. If one leg and the hypotenuse of a right triangle have lengths 1 and 2,
respectively, find the length of the other leg.
17. Find the I ength of a leg of a right triangle if the other leg and the hypote
nuse have lengths 1 and \/3~, respectively.
18. Find the length of the leg of a rigfrl triangle if the other leg and the hy
potenuse have lengths 100 and 100\/3~ t respectively.
19. Given A ABC with m£ C  90 and with D the midpoint of AS aiid£
the midpoint of EC, prove that ACED m ABED.
20. For AA BC t m Z C = 90 and D is the midpoint oiA~B. If AC = y7 and
BC = 3, find CD.
21. Complete the proof of Theorem 10.16 by showing that
ABC «— * ACD
and
ABC * — ► CBD
are similarities. (See Figure 1012.) It will then follow from the equiva
lence properties of similarity for triangles that ACD * — * CBD is also
a similarity.
22. Prove Corollary 10.16.2 fOT the leg AC in Figure 1012.
23. Prove Corollary 1Q.1&2 for the leg EC in Figure 1012.
24. See the proof of Theorem 10.17. Show that a 3 + fe* = c 2 . (See Figure
1013.}
25. If A ABC ~ A DEF and 5 • A = 3 * DE, what is the ratio of the length
of an altitude of the smaller triangle to the length of the corresponding
altitude of the larger triangle? Which theorem justifies your answer?
26. In Kxcrcise 25, what is the ratio of the area of the smaller triangle to the
area of the larger triangle? Which theorem justifies your answer?
10.8 Some Right Triangle Theorems 429
27« If 16 \APQR\ = 25 ■ A ABC and If APQR — AAHG, what is the
ratio of PR to AC? Which theorem justifies your answer?
28. Prove Theorem 10.14 for the case in which D = B or D = C
29, Prove Theorem 10.14 for the case in which DBC (the proof for the
case in which BCD Is similar to the proof for the case in which
DBC).
10.8 SOME RIGHT TRIANGLE THEOREMS
Following arc some theorems regarding right triangles. Although
the) f are not profound, they arc useful theorems that every mathemat
ics student who has studied formal geometry ought to know. These
theorems should not surprise you. If you worked the exercises in Ex
ercises 10.7, you will recognize them as "old stuff."
THEOREM 10,19 The median to the hypotenuse of a right tri
angle is onehalf as long as the hypotenuse.
Proof; Let AABC be a right triangle
with D the midpoint of the hypotenuse*
{See Figure 1014.) We want to prove
that CD = \*AB, or equivalent! v, that
CD = DB. I^et point E be the midpoint
of CB. Then
C B
Figure 10*14
(BD, BE) = (BA, BC) (Why?) and LB^LB.
It follows from the S.A.S, Similarity Theorem that A ABC— A DUE.
Then
mLBED = mLBCA = 90 = mLCED.
It follows from the S.A.S. Congruence Postulate that ACDE=*
ABDE and therefore CD = DB.
There are some special right triangles that are referred to in special
ways. First we mention the 3, 4, 5 triangles. A triangle whose sides
have lengths 3, 4 t 5 is a right triangle. We know this since 3* + 4* m 5 2 <
(Arc we using the Pythagoreun Theorem when we muke tills em illu
sion, or are we using its converse?)
Given a distance function, there are infinitely many 3, 4, 5 right
triangles. Indeed, if A is any point in space (infinitely many choices
here) and if B is any point such that AB = 5 (infinitely many choices
here), there arc infinitely many possible points C so that A ABC is a
right triangle with AB = 5, BC = 3 t and CA = 4, and infinitely many
430 Similarity Chapter 10
possible points C such that AB = 5, BC = 4, and CA — 3, But there
are many, many more, not included among these, that are also fre
quently referred to as 3, 4, 5 right triangles as our next theorem
suggests.
THEOREM 10.2(1 (The 3, 4> S Theorem) If a: is any positive num
ber, then every triangle with side lengths 3x, 4x, 5.r is a right
triangle.
Proof: Let &ABC be a triangle with BC = 3, CA = 4, AB = 5. Let
x be any positive number. Let &A'B'C f be any triangle with B'C = 3x,
CA f =: 4s, A'B' = 5x. Then AA'B'C  AABC. (Which triangle
Similarity Theorem do we use in making this deduction?) Since A ABC
is a right triangle, it foDows that A A 'B'C is a right triangle, and this
completes the proof.
A triangle is called a 3, 4, 5 triangle if its sides arc of lengths 3, 4, 5
or if the lengths of its sides are proportional to 3, 4, 5, AH 3,4, 5 triangles
arc right triangles.
THEOREM 10.21 (The 5, 12, i.1 Theorem) If x is a positive num.
ber and if the lengths of the sides of a triangle are ox, 12x, and 13x,
then the triangle is a right triangle.
Proof: Assigned as an exercise.
A triangle is called a 5, 12, 13 triangle if the lengths of its sides are
proportional to 5, 12. 13.
THEOREM 10.22 (The /, J, \/2 Theorem) If the lengdis of the
sides of a triangle are proportional to 1. 1, y/S. then the triangle is
an isosceles right triangle.
Proof: Let the lengths of the sides of a triangle be o, fc, c and suppose
{«, h c) = (1, 1, Vl).
Then there is a positive number k such that a = k • 1, h = k • 1, and
Gssk*y/£. Then
a = b,
t& + 6 s = fe 2 + Jt= = 2fc 2 ,
c 2 = (k^f = 2&,
a 2 + fc 2 = cK
Therefore the triangle is an isosceles right triangle.
10,8 Some Right Triangle Theorems 431
A triangle is called a I, 1, V5 triangle, or a 45, 45, 90 triangle, if
the lengths of its sides are proportional to 1 , 1 , \/S,
THEOREM 10.23 (The 1, y/3, 2 Iheorem) If the lengths of the
sides of a triangle are proportiona] to 1 1 \/3, 2, then it is a right tri
angle with its shortest side half as long as its hypotenuse.
Proof: Assigned as an exercise.
A triangle is culled a 1, \/S, 2 triangle if the lengths of its sides are
proportional to 1, y/3 t 2,
A triangle is called a 30, 60, 90 triangle if the measures of it* acute
angles are 30 and 60.
THEOREM 10,24 A triangle is a 30, 60, 90 triangle if and only
if it is a 1, \/3, 2 triangle with the shortest side opposite the 30 de
gree angle.
Proof: Let A A BC be a 1 , \/3 f 2 triangle an d k a positive mm Vie r such
that AC = fc, BC = s/Sk, and AB = 2k. (See Figure 1015.)
Let D he the point on opp CA such that CD — k. Then
1. AABC=z ADBC 1. Why?
2. AB = DB = DA 2, Why?
3. mlABD + mlBDA + mlDAB = 180 3. Why?
4. mlABD = mlBDA = m/.DAB = 60 4. Why?
5. ml ABC = mlDBC 5. Why?
6. ml ABC + mlDBC = 60 6. Why?
7. ml. ABC =30 7. Why?
8. ml8AC = 60 8. Why?
9. mlACB = 90 9. Wiry?
Since L ABC is opposite the shortest side of A ABC, this completes the
"if* part of the proof.
432 Similarity
Chapter 10
Suppose next that A ABC is a 30, 60, 90 triangle. (See Figure
1016.) Let D be the unique point on opp CB such that CB = CD,
Figure 10lfl
Draw DA. Then
10. AABC s AADC
11. m/LCAB = mZ. CAD = 30
12. mLBAD as mZADB = mlDBA = GO
13. BA = AD = DB
14. BC = CD
15. BC + CD = BD
16. 2BC a AB
Let BC = fc. Then
17. AB = 2k
18. (AQ* + (BC) 2 = (AB)*
19. (Aqz + A' 2 = 4*2
20. (AC)* = 3Jfc*
21. AC= VSJfc
22. (BC,CA.AB) = (1, x/3',2)
10. Why?
11. Why?
12. Why?
13. Why?
14. Why?
15. Why?
16. Why?
17. Why?
18. Why?
19. Why?
20. Why?
21. Why?
22. Whv?
This shows that A ABC is a 1, \/5, 2 triangle and hence the "only if*
part of the proof is completed.
Note that in some of these names for special triangles the numbers
are side lengths (or numbers proportional to them), whereas in others
they are angle measures. There should be no confusion in regard to the
30, 60, 90 name and the 45 f 45, 90 name. Because 30, 60, 90 are not
the lengths of the sides of any triangle, and 45, 45, 90 are not the
lengths of the sides of any triangle. Which postulate justifies this
statement?
L0.8 Some Right Triangle Theorems 433
EXERCISES 10.8
L If AABC is a right triangle with ml C = 90, AC = 60, BC  80, and
with D the midpoint of AB. find CD.
2, (See Figure 1014) Z ACS and Z DEB are congruent angles in the sit
uation represented by this figure. Find several other pairs of congruent
angles. (Six more pairs would be rather good.)
3. In the proof of Theorem 10,19 we asserted that ACD£a ABDE.
Write a twocolumn proof for this deduction.
4* In a book on the history of mathematics find something about the rope
stretchers in ancient Egypt Explain the connection between rope
stretchers and right triangles,
5. A baseball diamond is u square whose sides are 90 ft. long. What is the
distance (to the nearest foot) between first and third base*?
6. The figure represents a cube whose six faces are 1 in. by 1 in. squares.
Using the Pythagorean Theorem twice, once on ABCD and once on
AABD, find AD, Express the answer exactly using a radical if neces
sary. (Why is LABD a right angle?)
7. A room in the shape of a rectangular box is 15 ft wide, 1 8 ft. long, and
8 ft. high. Find the distance to the nearest foot between one corner of
the floor and a diagonally opposite comer of the ceiling.
In Exercises 816, the lengths of the hypotenuse and one leg t>i' a right tri
angle are given. In each exercise, the triangle is a 1. l t \/2 triangle, or a
1, \/3» 2 triangle, or a 3, 4, 5 triangle, or a 5, 12, 13 triangle. Determine
which one.
8. 100,50
9. 100,60
10. 100. 50 VI
11, 100,80
12, ]00, 50\/3
13, NX), 92^
14. \/^,iv^
15. 145, 116
16. 63, 25
434 Similarity
Chapter 10
17.
I .et^ A ABC be a right triangle with the right angle at C The midpoint
of AB is the center of a circle which lies in the plane of the triangle and
which contains the points A and B. Does the point C lie inside of the
circle, on the circle, or outside of the circle? Why?
The figure suggests a point A on a high bluff above a level plane. If the
angle of elevation of the point A from the point C is n 90 degree angle
and if it is 200 ft. from C to B t wliat is the height BA to the nearest 10 ft,?
Assume that / CBA is a right angle.
::iC^S, :*>"" ~
19, The figure represents an observer A in an airplane 5000 ft. directly
above a point 8 on the ground. If J3 T C T D arc three colllnear points on
the ground and if m L ABC = 90, m L BAC = 45, m L CAD s 15, find
to the nearest 100 ft. the distance from C to D.
20. Find three positive integers, a t b t c such that \/a, \/' , » V^&re the
lengths of the sides of a right triangle. How many such triples of positive
integers arc there?
21. If rand y are any positive integers, distinct or not, show that y/x, \/y,
\fx + tj are the lengths of the sides of a right triangle.
22. In the figure is shown a right triangle A ABC with CD the altitude to
the hypotenuse. If AD = 1, DC = r, BD = y, show that * = y/y.
10,8 Some Right Triangle Theorems 435
23. In the figure below, ABCD is a parallelogram with AB = 76, AD m 50,
m£A = 30, and h the length of the altitude from D to AS, Find
\ABCD\.
24. A parallelogram has adjacent sides of lengths 22 and 14. If the measure
of one of its angles is 30. find the area of the parallelogram.
25. Find the area of a rhombus of side length 12 if one of its angles has a
measure of 60,
2& The measure of each base angle of an isosceles triangle is 30 and each
of the two congruent sides has length 24.
(a) How long is the base?
(b) What is the area of die triangle?
27. In the figure. ABCD is a trapezoid with AB \TJD, AD = BC = 20,
CD = 28, and ml A = mlH = GO. Find \ABCD\.
Use the figure to complete the proof that in a 30, fiO right triangle the
side Opposite the 30 degree angle is onehalf as long as the hypotenuse.
Proof: In the figure, ml A = 30, m L B = 60, and D is the midpoint
of AB. Show that A BCD i,s equilateral and that
BC = CD a 1 • AB.
2d. In a 30. 60 right triangle, the length of the hypotenuse is 8 V^
(a) Find the length of the shorter leg.
(b) Find the length of the longer leg.
(c) Find the area of the triangle.
30, Prove Theorem 10.21.
436 Similarity Chapter 10
31. Prove Theorem 10.23.
32. If u and c are positive integers such that u > v, and if A ? B t C arc points
such dial AC = 2av, BC = u*  p 2 , AB = u* + &, prove that &ARC
is a right triangle. (This exercise also appears in Chapter 9, but it is good
for a repeat appearance here.)
33. See Exercise 32, If u uiid t: are relatively prime positive integers (this
means tluit no positive integer except 1 divides both of them), if it
and o are not both odd, and if u > c, then it can be shown that the three
integers, u 2 — c 2 , 2uc„ i* 2 + v' £ , are relatively prime. If the lengths of the
sides of a light triangle are relatively prime positive integers, the tri
angle is called a primitive Pythagorean triangle and the triple of its
side lengths is called a primitive Pythagorean triple. Two examples of
primitive Pythagorean triples are '(3,4,5) and (5,12,13). Find five
more primitive Pythagorean triples,
34. challenge problem, TTie figure shows a right triangle A ABC with
CD the altitude to the hypotenuse.
If AC = p, BC = q t CD = f, prove that Jj + i = \.
CHAPTER SUMMARY
The central theme of this chapter is SIMILARITY. The concept of
similarity is based on our experiences with objects which have the same
shape. The relationship of lengths in one figure to the corresponding lengths
in a similar figure suggests the idea of a PROPORTIONALITY. In this chap
ter we studied the properties of proportionalities and we used them in de
veloping the geometry of similar pjlygons.
The key theorems of this chapter include THE TRIANGLE PROPOR
TIONALITY THEOREM. THE CONVERSE OF THE TRIANGLE PRO
PORTIONALITY THEOREM, THE &&& SIMILARITY THEOREM,
THE S.A.S, SIMILARITY THEOREM, THE A.A. SIMILARITY THE
OREM, THE PYTHAGOREAN THEOREM, and THE CONVERSE OF
THE PYTHAGOREAN THEOREM.
The chapter concludes with a study of special right triangles. A knowl
edge of these triangles will prove useful as you continue your study of
mathematics*
Review Exercises 437
REVIEW EXERCISES
In Exercises 11 0, complete the statement so that it will be a proportionality.
1.(5.12) = (35,0)
2. (1,2,3,4, 5) = (0,0 EJ. Q 15)
3. (25, 60, 65) = (Q], 12,0)
4. a io f 2i) = (a. mm)
5. (4. 10, 21) = ([3,6, HI)
6. (4, 10,21)= (0, 0,6)
7. (5000, ,3000, 1500) = (200, 0, 0)
MiOO,400,5W) = (II],m,10)
9. (27, 27,81) = (a [7], 3)
10. (357, 1309, 833) = ([?}, 11, [7J)
In Exercises 1 120, determine If the given statement is true or if it is false.
11. If x = y, then (5, x)  (5, y).
12. lfxyt=y, then (5, x) = (5, ij).
14If f = f'**>(3.3)f (*>!/>>
15. If^^^.aieniAx)^^/
16. If x* = a&. then (a> x) = (*, £>),
17. If x* = ab, then (*, a) = (b, x).
18. If («. fc) = (c, <f), then (a. h) = [d, c),
19. If (a, fc) = (c, a*), then (a, <:} = (fe, rf).
20. If acf = hc t then (n, &) = (c, 5).
21. If AABC is any triangle, then AABC — AABC. Which property of
an equivalence relation docs this illustrate: the Reflexive Property, the
Symmetric Property, or the Transitive Property?
22. If A ABC A DICE then &DEF ~~ A ABC. Which property of an
equivalence relation does this illustrate?
23. If AABC ADKF and ADEF — AG//I, then A ABC— ACHL
Which property of an equivalence relation does this illustrate?
24. State the Triangle Proportionality Theorem and show how it may be
used to prove the theorem regarding the lengths of segments formed by
two transversals cutting three or more coplanar and parallel lines.
25. State the three triangle Similarity Theorems.
438 Similarity Chapter 10
2& According to Theorem 10,9, if A ABC is any triangle and k is any posi
tive number, then there is a triangle &DEF such that ADEF= A ABC
with proportionality constant t. F.xplain how tills theorem was used in
proving the triangle Similarity Theorems,
27. Explain why there is an A. A, Similarity Theorem but no A.5.A. Simi
larity Theorem.
2S. Prove that the altitude to the hypotenuse of a right triangle determines
two triangles that are similar to each other.
29. Using similar triangles, prove the Pythagorean Theorem.
30. If A, B> C, D are points such that ADB, AD = BD = 25, AC = 30,
BC = 40, find CD,
31. If the lengths of two sides of a right triangle are 10 and 15, find the
length of the third side. (Two possibilities.)
32. Find the measures of the angles of a triangle if the lengths of its sides
are \/2~. V? 2 
33. Find the measures of the angles of a triangle if the lengths of its sides
are \/S, 2\/3~.3
34. We know that SO, 60, 90 cannot be the lengths of the sides of a triangle.
Which postulate justifies this statement?
■ In Exercises 3543, refer to Figure 1017 in which A ABC is a right triangle
with the right angle at C, CD is the altitude from C to the hypotenuse,
AD = x, DB = «, AB = c,BC  a, AC = h, and CD = h.
Figure 1017
35. A ACS — A ADC and A ACS ~ A[TJ. By the symmetric and prop
erties of similarity for triangles, A[T] — A(7J.
36. h 2 equals the product of x and T] (o, b t or tj).
37. h 2 equals the product of x and T {a, c, or tj),
38* a 2 equals the product of y and \T\ (k c > or x}>
39. If x = 16 and tj = 9, find h,a, and h.
40. If D is the midpoint of AS, then CD = [T] (in terms of c),
41. If m L A = 30 and c = 15, then a = [TJ.
42. If a s 10 and c = 20, then b = [U and m£A = [?].
43. If ml A ^ 45 and a = 12, then b = JT] and c = [?].
Review Exercis«f 439
44. Copy and complete: In a 30, 60 right triangle, the side opposite the 30
degree angle is [?] (in terms of the hypotenuse).
45. If a boy 5 ft, tall casts a shadow 2 ft. long, how high (to the nearest 10 ft.)
is a tree if its shadow is 73 ft. long? What assumptions did you make in
working this exercise?
48. Find the distance from C to A& if AC = 10, BC = 10 V5, AB = 20,
47. If the hypotenuse and a leg of a right triangle have lengths 241 and 230,
respectively, find the length of the other leg.
48. challenge froblem. Given rectangle ADEH and points B and
C on AD such that HA = AB = BC = CD = DE = 1, prove that
m/LEAD = mlEBD = mAECD.
Chapter
Joyce fi. Wtlsoii/Photo Researcher*
Coordinates
in a Plane
11.1 INTRODUCTION
In Chapter 3 we introduced the idea of a coordinate system on a
line. Recall that if F and Q are any two distinct points* then there is a
unique coordinate system on FQ with P as origin and Q as unit point.
Tims a coordinate system on u line is determined by choosing any two
distinct points on the line, one of them the origin and the other the \ I j ii I
point If, on line Z, F is the origin and Q is the unit point, then FQ is
called the unit segment for the coordinate system on / determined by
P and Q. A coordinate system on I is a onetoone correspondence be
tween the set of all points of I and the set of all real numbers. The
numbers associated with the points of /are called coordinates, and they
can be used to determine distances (in the system based on FQ as the
unit segment) between points on I.
In diis chapter we introduce the idea of a coordinate system in a
plane. In a plane, each point is associated with a pair of numbers, rather
than a single number. After proving some basic theorems concerning
a coordinate system in a plane, we develop some equations for a line.
We then show bow coordinates can be used to provide simpler proofs
of some geometric theorems.
4.42 Coordinates in a Plane
Ctiapter 11
H.2 A COORDINATE SYSTEM IN A PLANE
Suppose that a plane is given and, unless we specify otherwise, that
all sets of points under consideration are subsets of this plane. Suppose
further than a unit segment is given and that all distances are relative
to this unit segment unless otherwise indicated.
Let OX and OY be perpendicular lines in the plane intersecting
in the point O as shown in Figure 111. Let / and / be points on OX
and Vi t respectively, such that
Ol = OJ = 1.
There is a unique coordinate system on OX with origin O arid unit
point I, This is called the \ coordinate system, and the coordinate of a
point R of OX in this system is called the xeoordiiuiteor abscissa of R.
t Y
6
i
6 s 4 321
•1
/,
It
X
12 3 4 5
Figme 111
In Figure 111, the abscissa of R is 5. There is a unujue coordinate
system on OY with origin and unit point J, This is (railed the
yeoordinate system, and the coordinate of a point S of OY in this sys
tem is called the ycoordinate or ordinateof S. In Figure 111, the ordi
nate of S is —2. Name the coordinates of Tand V in Figure 11L Is it
necessary to specify the coordinate system in each case? Why?
The line OX is called the ■ mlnmrl OY is called the yaxiv Together
they are called the coordinate axes. Their point of intersection, O. is
called the origin and the plane is called the .vvplimc. Although we
usually represent a line with a segment and an arrowhead at each end,
it is common practice to represent an axis with a segment having an
arrowhead only on the end that "points in the positive direction." In
many of the figures in this book axes are represented in this way.
11.2 A Coordinate System in a Plan* 443
The projection of a point P on a line f is (1) P itself if F is on /, and
(2) the foot of the perpendicular from F to /if ? is not on /. (See Defini
tion 10.4.) Since the perpendicular segment from an externa! point to u
Line is unique, each point in a plane has a unique projection on a given
line in that plane.
In Figure 112, PA is perpendicular to the xaxis at A and FB is
perpendicular to the {/axis at B. Therefore A is the projection of Fori
the xaxis and B is the projection of P on the (/axis. If 4 and 3 are the
coordinates of A and B> respectively, then we call the ordered pah* of
numbers (4, 3) the coordinates of F.
.
^
s
4
*
*
a
i
^
±A s
P5432:
I 2 i 'A 5

:
.
L.J
Figure 11S
More generally, if F is any point in the xyplanc, the xcoordinatc
(abscissa) of F is the xcoordinate of the projection of F on the xaxis.
The ^coordinate (ordinate) of F is the (/coordinate of the projection
of F on the {/axis. We call the xcoordinate of F and the incoordinate
of F the coordinates of P. The xycoordinates or, simply, the coordi
nates of F are an ordered pair of real numbers in which the abscissa is
the first number of the pair and the ordinate is the second. Thus if the
abscissa of F is a and the ordinate of F is 6, then the x {/coordinates of
P are written as (a, b).
THEOREM 11,1 The correspondence which matches each point
in the .riyplane with its xycoordinates is a onetoone correspond
ence between the set of all ordered pairs of real numbers and the
set of all points in the xyplane.
Proof: In the xt/plane, let an xcoordinate system on the xaxis and a
{/coordinate system on the {/axis be given. Let F be any point in the
given aryplane, If A and /? are the projections of f on the xaxis and the
44* Coordinates in a Plane Chapter 11
(faxis, respectively, let the abscissa of A be a and the ordinate of B f*?
k Since each point on the *axis has a unique xcoordinate and each
point on the yaxis has a unique (/coordinate, and since the projections
of P on the x and (/axes are unique, it follows thai there is exactly one
ordered pair of real numbers (a, h) that corresponds to the point P.
Conversely, let (a, b) be any ordered pair of real numbers; then
tiiorc is a unique point A on the xaxis with abscissa a and a unique
point B on the (/axis with ordinate h Also, there is a unique line fa
through A and perpendicular to the xaxis and a unique line fe through
B and perpendicular to the (/axis. These two Hues intersect (Why?)
in a unique point P. Hence every ordered pair of real numbers corre
sponds to exactly one point in the given xt/plane and the proof is
complete.
Definition 11.1 The onetoone correspondence between
the set of all points in an xi/plane and the set of all ordered
pairs of real numbers in which each point P in the plane cor
responds to the ordered pair [a, b), in which a is the xcoordi
nate of P and b is the ycoordinate of F, is an v l i: : mate
system.
Since there are many pairs of perpendicular lines in a plane and
since any pair of such lines may serve as axes, it follows that there are
many ^coordinate systems in a given plane. In a given problem sit
uation, we arc free to choose whichever coordinate system seems most
appropriate.
In view of the onetoone correspondence between the set of alt
ordered pairs of real numbers and the set of all points in a given
xf/plane, it is clear that symbols used to denote the ordered pairs may
be used to denote the corresponding points. Thus, if P is the point
whose coordinates are the ordered pair (4. — 3) t we may speak of the
point (4, —3) or we may write F— (4, —3). Sometimes we simply
write P{4, 3).
It should be noted that the numbers in an ordered pair need not
be distinct. Thxis (3, 3) is an ordered pair of real numbers, Of course,
the ordered pair (3, 5) is not the same as the ordered pair (5, 3). Indeed,
(0, b) = {e. d ) if and only if a = c and h = d.
As shown in Figure 1 13, it is customary to think of the unit point 1
as lying to the "right" of the origin (that is, on ray OX in the figure) and
of unit point 1 as lying "above" the origin (that is, on ray OY). This
means, then, tin at the points on the xaxis with positive abscissas lie to
the right of the origin and those points on the xaxfs with negative
1 1*2 A Coordinate System In a Plant 445
A
'V
rS
*?
x i
3
5
ngmm 1 t<3
abscissas lie to the "left" of the origin (that is., on opp OX). Where do
the points on the yaxis with positive ordinate* lie? Where do the points
on the yaxis with negative ordiriates lie?
There are situations in which it is convenient to think of the posi
tive part of the iaxis as extending to the left, or the positive part of
the yaxis as extending downward, or some other variation. However,
it will not be necessary to do this.
A line in a plane separates the points of the plane not on the line
into two half planes, the line being the edge of each halfplane. Simi
larly, the coordinate axes separate the points of an xyplune not on the
axes into four "quarterplanes," or quadrants, the union of whose edges
is the axes. For convenience, these quadrants are numbered 1, 1 J, ITT, IV
as indicated in Figure 1 1 4 .
+y
a
*<0
y>0
x>Q
y>0
x \
in
IV
x<0
*>0
>'<0
Figure lii
Quadrant I is the set of all points {x t tj) such that % > and y > 0.
Quadrant II is the set of all points {x t y) such that x < and y > 0,
Quadrant III is the set of all points {x, y) such that X < and y < 0,
Quadrant IV is the set of all points (r. y) such that x > and y < 0.
446 Coordinates in a Plane Chapter 11
We can describe the coordinates of those points (x, y) on OX by
x > and y = 0, and the coordinates of those points on opp OX by
x < and y = 0. Describe, in a similar way, the coordinates of those
points (i% y) on OY: on opp OY,
Since we usually think of an xi/eoordinate system oriented as we
have shown in Figures 111 through 11^1, it is customary to call all lines
parallel to OY vertical lines. Similarly, we cal all lines parallel to OX
horizontal lines.
It is often convenient to use **above," "below," "right," "left" to
describe the position of a point. However, we can get along without
these words if challenged to do so. For example, we could describe the
position of the point P = (2, —5) by saying that P is 5 units "below"
tlie xaxis and 2 units to the "right" of the yaxis in an xiypl&ne. Or we
could say that V is in the fourth quadrant, that it is on a vortical line
which intersects the xaxis in a point 2 units from the origin, and diat
it is on a horizontal line which intersects the yaxis in a point 5 units
from the origin.
EXERCISES 11.2
■ In Exercises 1S, name the quadrant in which the paint lies.
h (2,4) 5. {*, 2)
1 (7, 3) 6. (7.3, 1)
& (V2, 5) 7. (V5, V3)
•I)
4. (1.2, 6) 8. & I)
In Exercises 918, describe the set of all points (x, y) which satisfy the given
conditions.
9. x <C 0, y < 14. x is any real number, y = 4
10. x > 0, y < 15. x = —2, y is any real number
1L x < 0, y = 16. x > 3 S y = 5
12. x = 0, y > 17. x = 2, y < 1
13. x = 0, y < 18. xy =
19. If it = {3, 7), and
(a) if S is the point where the vertical line through H intersects the
xaxis. what is the abscissa of S? The ordinate of S? What are the
coordinates of S?
(I) if T is the point whore ihc horizontal line through /{ intersects the
yaxis, what is the abscissa of 7? The ordinate of T? What are the
coordinates of 2*?
11.3 Graphs In a Plane 447
20. Of the following points, 8nd three that are eollinear: (3, 5), (5* 7),
(—5, —5), .5, 2), (ff, —5)*
21. Describe the set of all points in the xj/*plaiie for which the abscissa is
—2; for which the ordinate is 6. Describe the intersection of these two
set\
22. Describe the set of all points in the xi/plane for which the abscissa is
zero; the ordinate is zero, Describe the intersection of these two sets.
Describe the union of these two sets.
11.3 GRAPHS IN A PLANE
A graph is a set of points. To draw a graph or to plot a graph is to
draw a picture that suggests which points belong to the graph. The
picture of a graph shows the axes, but they are not usually a part of the
graph. Of course, a subset of the axes is often a part of a graph.
It is customary to label the x and yaxes as shown in Figure 1 15.
It is usually desirable to label at least one point (other tlian the origin)
on the xaxis with its xcoordinate and at least one point (other Ulan the
origin) on the yaxis with its (/coordinate.
*
4
3
2
:
■
* fc
43
1 2 3 4
_1
Li
Figure 115
In setting up an xycoordinate system we start with three distinct
points O, /, / such that 57 X 0?and Ot =_0/ = 1. as in Figure 111.
Although it is understood that Ol and OJ arc congruent segments
(based on lengths in the distance system that we consider to be fixed),
it is sometimes helpful, particularly in applied problems, to take points
T and / so that IT! and OJ "appear" to be different in length. If this is
done, a picture of the xt/plane may be described by saying that the
"scale" on the xaxis is different from the "scale" on the yaxis. The
word scale, as used here, is not part of gilt formal geometry. As far as
our formal geometry is concerned the distance from O to I is the same
as the distance from O to / regardless of appearance.
448 Coordinates in a Plane
Chapter 1 1
If the scale on the xaxis is different from the scale on the yaxis, a
graph may appeal 1 distorted. An answer to the question "When is a
square not a square?" might be "When different scales for the & and
yaxes are used in graphing its length and width/' For example, Figure
il6 shows a picture of a quadrilateral ABCD all of whose angles are
right angles. Since AB = CD = 4 and AD a BC = 4 in our formal
geometry, it is true that ABCD is a square in our formal geometry. In
physical (Informal) geometry, if we were to measure the sides of quad
rilateral ABCD with a ruler, wo would find that Ihey are of unequal
length and conclude diat ABCD is not a square .
Figure 116
t
*v
1
—
12
WIO
cm 10)
■
r
T
1
4
Ai2 t 6)
fltf.G)
^
11
12 3 4 5 6
1 ! 11
If a graph contains only a few points, it may be desirable to write
the coordinates of each point beside the dot that represents iL
Example 1 Plot the points A(4, 3), B(0, 5), C{5, 2), D(4, 0).
(Sec Figure 117,)
*y
1 7]"
*i
4,3
)
0(4,0)
*l
3
Li
3
I'll
2)
,5
9
%fy
.
Hxwe 117
Sometimes a small open circle is used to indicate that the endpoint
of a segment or of a ray does not belong to the graph. In this connec
tion recall the symbol for a halflme introduced in Chapter 2.
11.3 Graphs in a Plane 449
Example 2 Draw die graph of {x t y) ; x < 2 and y = 3}. (Sec Figure
4
'V
!
2
x k.
;
J
3
r
Figure 11S
If there are infinitely many points in the graph, the picture may
contain segments or curves, and sometimes shaded regions or arrows,
to indicate which points belong to the graph.
Example 3 Draw the graph of {fr y) : 1 < x < 3 or 2 < y < 4}.
(See Figure 119.)
>5
■IE IE
H t
f
Urn , fc u
«.
L :
«' — 
if
— — " "* m 2
1
i
5
I *
% "
, .
i 1 1
Figure 119
Note that in Figure 119 we have shown l± and h as solid lines to
indicate that they are a part of the graph. We have shown parts of mi
and J7»2 as dashed lines to indicate that these parts do not belong to the
graph. Of course, segment AD on mj and segment BC on m 2 are part
of the graph. It is desirable to indicate all of lines vi] and m% in some
manner since they are a part of the "Iwundary" of the graph. Let us
agree, then, that if a line, a segment, or a ray is not part of a graph, but
serves as a boundary to a graph , it will be shown in the graph as a
dashed line, segment, or ray.
Write the coordinates of the point of intersection of lines 1% and mi
in Figure 119. Is this point a point of the graph?
450 Coordinates in a Plane
Example 4 Daw the
Chapter 11
of {{x, f) :*< 2}. (See Figure 1110.)
Figure 1110
As Figure 1110 suggests, the graph of {(*, y) : x < 2} is the
half plane on the left of the vertical line /. The interiors of the rays
shown in the graph are intended to indicate this halfplane. Why does
line I appear as a dashed line in this graph? Since line I is a vertical line,
it is parallel to the yaxis. Therefore all the points of line I are on the
same side of the yaxis in the xyplane. Another way of describing the
graph in Figure 1110, then, is as the halfplane with edge / on the op
posite side of line I from the yaxfc. Since line / represents the set of all
points hi the ,ri/plane whose abscissa is — 2, another way of describing
line / is the line with an equation x = — 2, The halfplane pictured in
Figure 1110 is the set of all points (x, y) such that x < — 2; or we may
describe it as all of the *r/plane which lies to the left of the tine
x= 2.
EXERCISES 11.3
1. If A = (2, 2), B = (3, 3), C = (4, 2), D = (3, 3), draw the
graph of the set of all points winch belong to the polygon A BCD.
2. Draw the graph of the set of points that belong to the interior of the
polygon in Exercise 1 .
3. In your graph for Exercise 2, should the segments 7iB, T5C, CD, DA~ ap
pear as dashed lines or as solid lines?
4. Draw the graph of
{(x, y) : x = 3 and — 1 < y < 2}.
5. Is the point (3. 2) part of the graph for Exercise 4? Is the point (3, — 1)
part of the graph?
11,3 Graphs in * Plan* 451
hi Kxercises 13, trvsph lllfi led of all pimiK ;>', (/) in un Xgf pkne satisfying
die given conditions. Then describe the graph of the set in words. Exercise
6 has been worked as a sample.
6. 2 < x < fi and y = 3.
Graph:
'
*
it&3j 5(6,3)
2
x t
2
2
n
1
Figure 1111
7.
10.
1L
12.
13.
R
15.
16.
17.
18.
20.
Dcscriptiou: The graph is the segment whose endpotnts are
A{2, 3) and B(0, 3).
x > 2 and y = 2
x = 5 and i/ >
*>3
K = 3 and 2 < r/ < 5
—5 <x< — lor2<i/<5
— 5 < x < — 1 and 2 < y < 5
Which, if any, of the following points ure not part of the graph for Ex
ercise 12: ( — 5. 2), ( — 1, 2), ( — 1, 5), ( — 5, 5)? Which of these points are
not part of the graph for Exercise 13?
If P = (2. 0) and Q = (9. 0), what is the length of segment ¥Q? Justify
your conclusion.
If A = (2, 5) and B a (8, 5), what is the length of ATJ? Why?
If C = { 3,  4) and D = ( 3, 6). what is the length of CD? Why?
Give the ctHirdinates of the midpoints of the segments whose endpoints
are the following:
(a) (3, 2), (3. 12)
(b)(M>,(9.4)
(c) (3,1), (7,1)
chaukn'CE probiem. Give the coordinates of the midpoint of the
segment whose endpoints are (3 T 5) and (8, 7).
challenge pkoblem. If F = (5, 1) and Q = (8, 6), what is the exact
length of FQ?
452
Coordinates m a Plane
Chapter 1 1
11.4 DISTANCE FORMULAS
Consider an ^coordinate system as shown in Figure 1 J 12. Let J
be the unit point on the itaxis and let P and Q be any two points with
al jseissas % l and x 2 , respectively, on the .taxis.
'*>y
i £ 9 x
«1 3 x 2
r^giire IMS
We know, by the definition of a coordinate system on OX relative to
unit segment Ol (Definition 3.3), that
PQ (in 01 unite) = s,  x 2 \ = (**  «,.
For example, if P = (3, 0) and Q = {7, 0% then %i = 3, x* = 7, and
PQ = 3  7 = 7  3J = 4.
Now suppose that I is any line in the xyplanc and parallel to the
.Tajds as in Figure 1113 (that is, I is any horizontal line), Let P and Q
be any two points on I and let Pi(x u 0) and Qi{x 2 , 0) be the projections
of P and Q t respectively, on the xaxis.
Q
3— t*
f,
0:
■i":
i 2
Figure 1113
By our definition of the abscissa of a point in an ryplane, we know that
the abscissa of Pisxj and that the abscissa of Q is x 2t If P = Pi and
Q = Q u then
PQ = PyQi = fa  X2J.
11,4 W>tar»ce Formula* 453
If P=£P lf then Q=£Qi and the quadrilateral PPiQi() is a parallelo
gram. Therefore we again have
PQ = P t Q, = \x,  x 2 \.
Note that iflisa horizontal line, then / is perpendicular to the yaxis
(Why?) and every point of I projects onto the same point in the i/axis.
Thus all points on a horizontal line have the same ordinate. We have
proved the following theorem.
TJJEOBE \f 11.2 If p( Xit y T ) and Q(x 2 * ij\) are points on the same
horizontal line in an xi^plane, men
PQ = *i  *!.
It should he noted that if P = Q in Theorem 11.2. then
x, a Xt and PQ = = 0.
TllEOR EM 11.3 UP(x l ,y 1 ) and Q(x if f/ 2 ) are points on the same
vertical line in an .viyplaiie, then
pq  m  y«l
Proof: The proof is similar to the one given for Theorem 11.2 and is
assigned as an exercise.
Example I If A = (3, 5) and B = (6, 5), find AB.
Solution! AB = [3  (6) = 9,
Note that if two points have the same ordinate, then they lie on the
same horizontal line. Thus A 6 in Example 1 is a horizontal line. Hence
AB is die absolute value of the difference of the abscissas of the points
A and B t as shown in the solution to Example 1.
Example 2 If C = (2, 3} and D = (2, 4), find CD.
Solution: CD m 3 4 = 7 = 7.
Note that if two points have the same abscissa, then they lie on the
same vertical line. Thus CD in Example 2 is a vertical line. Hence CD
is the absolute value of the difference of the ordf nates of the points C
and D, as shown in the solution to Example 2.
4W Coordinates in a PJarre
Chapter 1 1
You have seen how to find the distance between any two points in
an aycoordinate system if the two points lie on the same horizontal line
or if the two points lie on the same vertical line. Next we show how to
find the distance between two points if Lhey lie on the same oblique
line, that is, a line that is neither horizontal nor vertical Before pro
ceeding to the general case, let us consider an example.
Exampk3 UP = (2, 3} and Q = (4, 5), find FQ.
Solution: {See Figure ! t  14.) Let h be the line through Q and parallel
to the yaxis and let h be the line through F and parallel to the xaxis.
These twojines intersect in a point R such that h ± h at R, Why?
Therefore FQ is the hypotenuse of a right triangle, &PQR, with the
right angle at R, Since f 2 is a horizontal line and l± is a vertical line, we
have
PR = 4  {2} = 8
QR = 5  (3)1 = 8
by Theorems 11.2 and 11.3.
By the Pythagorean Theorem,
to 2 = m? + (<?w,
= 62 + 82,
= 36 + 64,
= 100,
and hence
PQ = 10,
,55
t*i
i
A mm
/ i
 /
/
'
' \
A
! *
i
4
~H
_ ld__ k
2,3)
]R<A,3) ~^
1 1
I
i
Figure 1114
Note that in finding PQ in Example 3 we made reference to a right
triangle. We proceed now to the theorem that, will enable us to find the
distance between any two points in any ^coordinate system without
reference to a right triangle. The formula in this theorem is often re
ferred to as the Distance Formula for point 1 ; in an xuplane.
THEOREM HA If F, = (*,, pj and P 2 = (x 2t y 2 ) are any two
points in an xyplane s then
PiPi = V(*T *a) a + (m  y 2 ) 2 .
Proof: There are four cases to consider.
Cam L Fi = F 2 .
Case 2. F L and F 3 are distinct points on a horizontal line.
Case 3. Fi and P 2 are distinct points on a vertical line.
Case 4. Pi and F 2 are distinct points on an oblique line.
11,4 Distance Formuttt 455
Proof of Case 1: If Pi = P** then x± = xj, yi = yz y and, by the for
mula of Theorem 11.4, we have
P& = Vo = o,
as it should since the distance between a point and itself is defined to
be zero.
Proof of Case 2: If P t and ?2 axe distinct points on the same hori
zontal line, dien y± = tj2 and, by the formula of Theorem 1 1 .4, we have
PlP2 = \Z(X!  X 2 )2 = h  X 2 .
This result agrees with the statement of Theorem 11.2 for two points
on the same horizontal line.
Proof of Case 3: If Pi and P 2 are distinct points on the same vertical
line, then Xi = x% and, by the formula of Theorem 11.4, we have
PiC«i.ja
P1P2 = y/(y L  1/2) 2 = yi  U2I
which agrees with the statement of Theorem 11,3.
Proof of Case 4: If P t and P 2 are distinct points on an oblique line as
shown in Figure 1115, then the line through Pi and parallel to the
y2Lsis intersects the line through P% and parallel to the xaxis in a point
H(xy, r/2) such that APiP^H is a right triangle,
P 2 R = [xi  %z\
by Theorem 112 and
Piil = Ij/j.  y a \
by Theorem 1 1 .3. We have
(p 2 Rr = ,*,  x 2 p = (x, 
and
Since PiPz is the hypotenuse of AP1P2R, we have, by the Pytha
gorean Theorem, that
J^*2.j'a)
or that
PxP* ~ V
and the proof is complete.
(P1P2) 2 = (P 2 H) 2 + (PxH) £
fi  X2P + (#,  ya) 5 .
456 Coordinates in a Plane Chapter 11
Since (xi  $$* = (x 2 — *i) 2 and (y t  y 2 ) 2 m (y 2  ifi) 2 . in
using the Distance Formula of Theorem 11.4, it does not matter which
point is designated P t (x lt tji) and which point is designated Pg(* 2 » y 2 )>
Example 1 If A = (2, 4) and B = (5, 3), find AB>
Solution: Substituting the coordinates of the given points in the Dis
tance Formula, we have
AB=W (S)p + (4  3)4
= yya + (_ 7) 2
= y/49 + 49
1
= 7\/2
Note that in working Example 4 we considered A(2, — 4) M the
point J*i(xi, 1/1) and B( — 5, 3) as the point Pafet Jte) when we substi
tuted these coordinates into the distance formula. Show, hv working
Example 4 again, that we would have obtained the same result for AB
had we considered A(2, — 4) as the point Pz(x 2 , y 2 ) and B( — 5, 3) as the
point Pxfa, ^i).
EXERCISES 11.4
1. Prove Theorem 1].3.
In Exercises 210, use Theorem 1 1 .2, Theorem 1 1 ,3, or the distance formula
to find the distance between the given points.
2. (3, 2) and (3, 11)
3. (f I© and <& ID
4. (2.5, V^ and (17.3, y5)
3. [w, 4.8} and fa 9.6)
6. (2, 7) and (3, 5)
7. (3. 6) and (9, 0)
8. (4 r 17) and (3,9)
9. (3, 5) and (5, 1)
10, (6, 3) and (4, 2)
IL Find the perimeter of die triangle whose vertices arc A( — 2, —3),
B(3, 9), and C( 10,12).
1L4 Distance Formula* 457
12. Prove that the triangle whose vertices Are P{1> 2), Q(9 t 2), and /l(5, 8)
is isosceles.
13. AAHC lias vertices A (6. 0), B(4 t 4), and C(10, 4).
(a) Find the perimeter of A ABC.
(b) Find the area of AABC.
14. Find the lengths of the diagonals of a quadrilateral ABCD if
A = (4, 3), B = (7, 10), C = (8, 2), and D = (1, 5}.
15. The vertices of APQR are P( 1, 2), ()(4, 0), and fl(2, 5). Prove that
A PQR is a right isosceles triangle.
16* If the distance between A(B t 2) and B{0 t tj) is 10, find the possible
tf^oordtnates of B*
17* Find the coordinates of the points on the xaxis whose distance from
(2, 8) is 10.
IS. If (a, — a) is a point in quadrant IV, prove that the triangle with vertices
(—5. 0) f (0, 5), and (o T — a) is isosceles.
19. Given D = (2, 2), E = (10, 2}, F = (4, y) with Z DKE a right angle,
find the two distinct possible values of jf.
20. Given P = (2, 7), Q = (3, 3), R = (6, 9), use the distance formula
to show that PQ + QR = PR, and hence that FQR.
■ Exercises 2126 refer to the triangle whose vertices are A = (2, 4), B =
(6, 8), and C = (12, 2).
21. Draw line /i through A and parallel to the r/axis. Draw line / 2 through
B, parallel to the xaxis, and intersecting f t at D. What are the coordi
nates of D?
22. Draw line h through C, parallel to the xaxis, and intersecting h at E.
What arc the coordinates of £?
23. What kind of quadrilateral is quadrilateral BDEC? Find \BDEC .
24. What kind of triangles are ABDA and ACEA? Find \ABDA\ and
AGFA .
25. Copy and complete:
BDEQ m \AABC\ + \A[T\\ + \&JB ,
26. Find \AABC\.
27. challenge problem. Find \APQR if P = (6, 0), Q = (1, 5), and
R = (10, 8).
2& CHAU.EXCE problem. If A = (0, 0), D = (b t c), B = {«, 0), and C =
(a ■+■ b* c), where a, b t c are positive numbers, prove that quadrilateral
ABCD is a parallelogram. (Hint If the opposite sides of a quadrilateral
are congruent, then the quadrilateral is a parallelogram.)
4M
Coordinates in a Plane
Chapter 11
11.5 THE MIDPOINT FORMULA
We know that the midpoint M of a segment PQ is the point be
tween P and Q such that PM = MQ or thai PM = JPQ, or, similarly,
M@ = ^P(X Suppose that the coordinates of P and Q arc given and
that we wish to find the coordinates of M, the midpoint of J 5 ^. For ex
ample, if P = (2, 3) and Q = (8, 5), we can find the coordinates (# t/)
of M, the midpoint of FQ t as follows. (See Figure 1116.)
'■'
1
aotn
t
M"i0.y)
Mix:
■<*
£r
l«*3>
1
!
i
i i
X
^(2,0) M'{x,0) Q'£8,0)
 1
! !
Ffcm Hie
Let F(2 T 0), AT(x, 0) t (X(8, 0) be the projections of P, M, Q, re
spectively, on the *axis and F'(0, 3 )* W'{0, i/), Q"(0, 5) be the projec
tions of P, M, Q f respectively, on the yaxis. Since If is the midpoint
of FQ, it follows from Theore m i 10. S that M' is the midpoint of FQ'
and M" is the midpoint of Fp 77 . Therefore, by Definition 3,3,
2 < x < 8 and 3 < t/ < 5< By the definition of midpoint, we have
(1) FW = M'Q' and (2) V'M" s M'V ■
By Theorem 11.2,
FM' = \x  2 and Afp* = 8  *
Sinee x — 2 > (Why?) and 8  a: > 0, we get by substitution into (1),
x  2 = 8  x.
Therefore 2x = 10 and x = 5.
Similarly, by Theorem 11.3,
F'M" = \y3\ and M"Q' f = \5  y\.
Since y — 3 > and 5 — y > 0, we get by substitution into (2),
y  3 = 5  y t
Therefore 2y = 8 and y = 4 Hence the coordinates of .W, the mid
point of PQ, are (5, 4),
11.5 Th« Midpoint Formula 459
We can proceed, as in the preceding example, to find the coordi
nates of the midpoint of any segment if the coordinates of the end
points of the segment arcs given. However, Theorem .1 1.5 provides us
with a formula that will enable us to find the coordinates of the mid
point of a segment. The formula in Theorem 1 1.5 is often referred to as
file midpoint formula.
THFnREM 1L5 UP = {x ly yi) and Q = {x 2 , y 2 ) are any two dis
tinct points in an xy plane, then the midpoint M of FQ is the point
»= fr^>
There are three cases to consider.
Cose 1: P and Q are distinct points on a horizontal line.
Case 2: P and Q are distinct points on a vertical line.
Case 3: P and Q are distinct points on an oblique line.
We shall begin the proof for Case 3 and assign the remainder and
the proofs of Cases 1 and 2 as exercises,
Procifof Case 3: Let P(xi,y t ) and ()(X2, 1/2) be any two distinct points
on an oblique line in an syplane as shown in Figure 1117 and let
M(x, y) be the midpoint of FQ. Let F(xi, 0), M'(x, 0), and Q'{x 2 , ()) be
the projections of P> M, an d Q, r espectively, on the %axis. By Theorem
10.3, M' is the midpoint of r<^ r . By Definition 3.3, x is between xi and
X% dius
<*< x%
or
X'l < * < Xt<
(In Figure 11 17 we have shown x± < x < x 2 , but this order might be
reversed if P and Q are chosen in a different way.)
<jfa2,^)
F(*1,Q) M'lx,0) QCx 2l CI)
Figure IM7
460 Coordinates in a Plane Chapter II
By the definition of midpoint,
par = arg\
and by Theorem 115,
FAT = \x  %i\
and
M'Q f = ,**  4
Therefore
\X  Xi\ = t2  X\>
If *i < i < x 2 , then x — xi > and *a — x > and we have
x — *i = x 2 — x. Why? Therefore
2x = *t 4 x 3
and
" = —2—
If ^2 < x < xi, then xi — x > and x — x 2 > 0. In this case,
FM' = x  ri. = xi — x
MV = x 2  x\ = x  z 2 .
x — x 2 = Xi — x,
and again we have
_ * + »
2 '
We have shown that if M (x, y) is the midpoint of the segment whose
endpoints are P{x yy iji) and Q{x 2 , ys), then the abscissa of \f is * 2 .
In a similar way, it can be shown that the ordinate of M is ** 1 _ ^ 2 ,
(Yon arc asked to show this in the Exercises, thus completing the proof
of Case 3.)
hxamplc 1 Find the coordinates of the midpoint of the segment
whose endpoints are the following:
1. A(2. 3) and 5(2, 9).
2. C(12, 1) and D(3, 1).
3. E(2, 7) and F{W t 12),
11,5 Th* Midpoint Formula 461
Solution:
L Segment XB lies on a vertical line. Therefore i = x\ = *t and
*i + *£ *i + JCi
—IT = ~2— "*
Therefore the midpoint is
2. Segment CD lies on a horizontal line. Tlierefore y = t/i = 1/2
tfi + ya _ yi ■+■ tji __
2 "* 1 '
2
Therefore th«
j midpoint is
Ma
V 2 *
yi)
/12 + (
3)
.»)(*»>
3. Segment EPlies on an oblique line; hence the midpoint M of EF
is the point
Example 2 The vertices of a triangle are A(0, 5), B(4 f 3), and
Q2, 1). Find the length of the median to BC.
Solution; The median to BU is the segment whose endpoints are
A{Q t 5) and M, the midpoint of BC, The coordinates of M are (1, 2) by
the midpoint formula; hence
AM = y/(Q  If + f£  2) 6
= VTT9
= V10
by the Distance Formula.
462 Coordinates in a Plane Chapter 1 1
EXERCISES 11.5
1 . Complete the proof of Case 3 of Theorem 11.5 by showing that the ordi
nate of M is Hl±Jil , (See Figure 1117.)
IS
2. Prove Case I of Theorem 11.5. (In this case, y = tj\ = tj2 in the state
ment of the theorem.)
3» Prove Case 2 of Theorem 1 1 .5. (In this case, x = i] = r* in the state
ment of the theorem.)
■ In Exercises 410, find the midpoint of AB if A and B have the given
coordinates,
4. (5, 2) and (5, 6)
5. (3, 5) and (8,5)
6. (0, 0) and (8, 10}
7. (0,0) and (8, 10)
8. {1, 2) and (6, 14)
9. (r, 7) and (3r, 3)
10. fab)md(5a,7b)
11. The vertices of a triangle are I\2, 3), £{10, 1), and K(4 t 6). Find the
midpoint of FQ. Find the length of the median to FQ.
12. The vertices of a triangle are A(3, 2), B(l, 6), and C(5, 2), Find
the lengths of the medians lo A~B and BC« How are the lengths of these
two medians related to each other? What kind of triangle is A ABC?
■ Exercises 1315 refer to quadrilateral ABCD whose vertices are the points
A = (0, 0), B = (6, 0), C = {8, 4), D = (2, 4).
13. Find AC and BIX
14. Show that the midpoint of AC is the same point as the midpoint of B7X
15. What kind of quadrilateral is ABCD?
M E*ercises 1624 refer to A ABC whose vertices are the points A = (2, 0),
B = (12, 0), C = (7, 5\/3).
I K. Prove that A ABC is equilateral.
17. Find the coordinates of D, the midpoint of AB.
18. Find the coordinates of E, the midpoint of fflOL
19. Find the coordinates of F, the midpoint of AC.
20. Show that the lengths oi' the three medians of A ABC are equal.
21. Find DE and show that t>E = JAC.
11.5 The Midpoint Formula 463
22. Find EF and show that EF = JAB.
23. Find FD and show that FD = \BC.
24. Do the results of Exercises 2123 prove that ADEF is equilateral?
Exercises 2527 refer to A/SK whose vertices are the points / = (0, 0\
S = (6, 0), K = (0
25. What kind of triangle is A JSK?
26. Find the coordinates of M, the midp>int of 5K.
27. Prove that JM s ]$K,
In Exercises 2831, the coordinates of two points A, M are given. Find the
coordinates of the point B suck that M is the midpoint of 3uB,
28. A = (1,3),M={4,7)
29. A = (4, 7), Af = (L 3)
30. A = (1, 8), M = (0, 0) „
31. A = (6, 4), M = (3, 2}
Exercises 3236 refer to the segment AH whose endpoints arc A = (3, 2),
B= (11, 6) and the point Pi x> y) on AB such that AP = %AB.
32. Let A', P\ /T be the projections of A, J* B. respectively, on the .taxis.
33. What is the absciss u of F in Exercise 32? What is the abscissa of P?
34. Let A", P", B" be the projections of A , P, B t respectively, on the (/axis.
35. What is the ordinate of P" in Exercise 34? What is the ordinate of P?
Mi, What are the coordinated of PI
37. If A = (4, 2) and B = (6, 3), find the coordinates of the point P on
AB such that AF = JAR (ERnfc See Exercises 3236.)
38. challenge problem. Given positive numbers a and h and a right tri
angle with vertices at A = (0, b) t B s (a, 0), and C = (0, 0), find the
coordinates of Af, the midpoint of A~B t and show that CM = ^AB. Does
tills prove that for antj right triangle, the median to the hypotenuse is
onehalf as long as the hypotenuse?
39. challenge PROBLEM. Given positive numbers a, h, and u quadri
lateral ABCD with vertices at A = (0, 0), B = (a, 0), C = (a + b, c),
and D = (b, c),
(a) prove that ABCD is a parallelogram,
(b) prove that the diago nals AlT and ED bisect each other by showing
that the midpoint of AC is the same point as the midpoint of B~D.
464 Coordinates in a Plane
Chapter 11
11.6 PARAMETRIC LINEAR EQUATIONS
If we are given a line in an xjyptune, it is often desirable to find the
coordinates of points on that line. We know that two distinct points
determine a line. Suppose that we know the xi/coordinates of two dis
tinct points A and B on AB. We should be able to find the xiycoordi
nates of any other point P on AB provided, of eoursc 5 enough informa
tion is given to determine one and only one point P on AB.
Example I Let A = (1, 3), B = (4, 7), and suppose that S, Q, fl are
^ y v ^ fr
points on AB such that 5 is on AB, Q is on AB, and R is on opp AB,
[See Figure 1118.) Suppose, further, that
AS = $AB, AQ = 2AB, AB. = AB,
and we wish to find the coordinates of S, Q t and iL
4
iy
J*
J
7$
/
.
/ 1 1
*£:
6
Ml
X s
;i
/
B)
flfa
X
~ 3 di
b
/ ~ 3
Figure 1118
Let S = (x, y). To find x and y we might proceed as follows. Let A\
B', S f be the projections of A, B % S, respectively, on the raxis, and let
A", B" t S" be the projections of A, B, S, respectively, on the yaxis.
Using Theorem 10.8, we get
2 _
3 "
AS _ A'S' *— 1
AB " A'B' " 3
2
3 "
AS A"$" y  3
AB A"B" 4
and
and
.r = 3,
9 3
11.6 Parametric Linear Equations 465
Similarly, we can find that Q = (7, 11) and R  (2,  1). (Verify
these results by selling up appropriate equations and solving Ihcm.) If
Fwere any other point on AB such thai
AP=kAB,
where k is any positive number or zero, we could find the coordinates
of P by a computation similar to those above. Our objective, however,
is to derive an expression from which the coordinates of any point on
AB can be obtained by simple replacements.
In Chapter 3 we studied a coordinate system on a line. In this chap
ter we have already defined an xi/coordinate system in a plane, based
on two line coordinate systems like those you studied in Chapter 3,
Let us consider now a third line coordinate system, a coordinate
« — *
system on line AB with A as origin and B as unit point We call it the
^coordinate system on AB, It should be clear that we start with an
,tf/coordinate system based on a unit segment for distance. When we
speak of "the distance*' between two points in die .ryplane, we are
talking about the distance based on the same unit segment that is used
in setting up the x^coordinate system. For every different choice of
points A and Bona line fin the i^planc, there is a different k coordi
nate system on I with A as origin and B us unit point. The ^distance
between two points on / will usually be different from the ^(/distance
between those points.
The following table shows the ^coordinate, the {/coordinate, and
the ^coordinate of the points A, B, S, Q, B that were shown in Exam
ple L
Point
xCoordinate
yCoordinate
feCoordiiiatc
A
1
3
B
4
7
1
S
3
s§
I
Q
7
11
2
R
2
1
1
P
m
13
k
We shall derive the equations which show us how to compute the
x and {/coordinates of a point on AB in terms of its fc coordinate. First
we shall show that the ^coordinates of the points of I form a coordinate
system on I as do the ^coordinates.
.:i,i.
Coordinate* in a Plane
Chapter 11
Let / be any nonvertical line in an .vyplaue. (See Figure 1119.) Let
O(0, y ), fft yi), P(x, y) be points on l y and let O t l\ F be their re
speetive projections on the x~axis. It foflows from the Plane Separation
Postulate and the properties of parallel lines that (J r t t F have the
same betweenness relation as do their respective projections. For ex
ample, if OIF. then Ol'F.
Figure IMS
Let § denote the unique coordinate system on I with O as origin
and I as unit point. Note that Ul serves as a unit segment for § and that
generally (except if / is parallel to the Yaxis) distances "in g" arc dif
ferent from distances "in the xtfcoordinate system.*' Let p be the co
ordinate of F in the system §, Since betweenness foT points on I is the
same as for their projections on the xaxis, it follows from the definition
of a coordinate system on a line that p and x are both positive, or both
negative, or both zero It follows from Theorem 10.8 that
OP
O'F
or '
Since these ratios are equal regardless of the distance function, we have
OP
OI
;i _ 1 ir " or "10
= Jx, and p as x.
Now S is the onetoone correspondence that matches each point P on
/ with a number p. Since p = x, we see that the correspondence that
matches each point of I with its xcoordinate is indeed a coordinate
system. Similady if lis a non horizontal line, then the correspondence
that matches each point of I with its yccKMdinate is a coordinate system
on I. We state these results in our next theorern.
TlfEQREM 11.6 If I is any nonvertical (nonhori/ontal) line in
an xiyplanc, then the onetoone correspondence between the
points of / and their xcoordinates ((/coordinates) is a coordinate
system on I.
11.6 Parametric Linear Equations 467
In the next theorem, as well as in many others throughout the rest
of this book, there is the assertion, 'A is real" within a setbuilder sym
bol. Ttiis is short for "k is a real number/'
THEOREM IL7 If A(x ls yi ) and B(x 2t y 2 ) are any two distinct
points, then
AB = {(x y) : x = x, + % 2  XjJ.y = {/! + %a  J/i),fcisreal},
If fc is a real number and if P is the point (x t y) where
x = *i + fc(x 2  «!) and ^ = yi + k(y 2  fiX then
fe a
AT
and
P€ A§
if
fc> 0;
"*4l and P£oppA$ if Ar < 0.
JVoo/: Let Aixi, y t ) t B{x 2 , f&) be any two distinct points. Suppose first
that AB is neither vertical nor horizontal, Think of three coordinate
systems on AB as suggested in Figure 1 120, the xeoordinate system
and the ycoordinate system, determined by the xycoordinatc system
(see Theorem 11.6), and the ^coordinate system in which A is the
origin and B is the unit point.
Figure 11*20
or
or
468 Coordinates in a Plane Chopttr 11
It follows from tlic TwoCoordinateSystems Theorem (Theorem
3.6) and its corollary, each applied twice, that AB is the set of all points
P(x 3 r/) such that
x — xi k — y — iji _ k —
x2 — xx 1 — * y2 — yi 1 — *
x  xi = *(* 2  Xi\ y  i/! = % 2  yt);
x = X! + *<*2  *i). y = yi + %2  yi);
where ^j^.
Since the definition of a coordinate system requires that hetween
ness for points agree with betweenness for coordinates, it follows that
P is on a5 if and only if h ;> 0, and ? is on opp AB if and only if k < 0,
If A E is a vertical line, then the xcoordinate of every point P on AB
is die same number, and there is no xeoordinate system on Ad, In this
case xi = xo = x for every point P{x t y) on AB and the equation
x = *! + *(* 2  *i)*
which simplifies to i = ii, is still applicable. The proof for the noii
vertiealnonhorizontal case is applicable to this case as far as the rela
tion between y and k is concerned. Therefore the assertion of the the
orem applies in the vertical case. Similarly, the theorem may be proved
for the horizontal case.
Tn
S3 = {(x, y) : x = x t + k(x 2  %§}, y = y± + k(y 2  yi), k is real},
the "it is real" is put in because k is not mentioned before the braces
and not before the colon within the braces, and we read the sentence as
"AB is the set of all points (x t y) such that
x = X! + k(x 2  x a ),
y = yi + %2  yi),
where k is real/'
The equations
X = x l + k(xi  Xi) and y = i/i + %z  f/i)
hi the statement of Theorem 11.7 are called parametric equations for
die line AB and k is called the parameter. A parameter j s usually
11.6 Parametric LI near Equations 469
thought of as a variable to which values may be assigned arbitrarily,
and other variables are defined in terms of it. By assigning real number
values to the parameter k in the parametric equations for AT?, we ob
tain ordered pairs of numbers corresponding to points on AB in the xy
eoordinate system. Of course, we could assign values to x (in the non
vcrtieal case) and find values of y so that the resulting ordered pairs
correspond to points of AB or we could assign values to y (in the non
horizontal case) and End values of x so that the resulting ordered pairs
correspond to points of AB. But it is easier, and it works in all cases, to
assign values to k and compute values of x and of y so dial the resulting
ordered pairs correspond to points of AB.
Note that parametric equations for a line aie not unique. If / is a
line, then there are many choices for the two points A and B of The
orem 11.7, and hence many pairs of parametric equations for L
If we replace the coordinates (% x 3 ) and (yj, j/g) in the parametric
equations of Theorem 1 1 .7 with the coordinates (1, 3) and (4, 7) of the
points A and B in Example 1, we obtain the parametric equations
x = 1 + 3k and y = 3 I 4k
for the line AB in that example. Recall that, in Example 1, S is a point
on AB such that AS = AB. Thus, by replacing k with the number f
In the parametric equations
X m 1 + m, y = 3 + 4Jt,
we obtain the coordinates (3, 5j) of the point S. Show that you get the
xycoordinates of the points Q and R on X5 in Example 1 by assigning
the values 2 and — 1, respectively, to k in the parametric equations
x = 1 + 3fc, y sb 3 + 4k.
It follows from Theorem 1 1,7 that every line can be represented by
setbuilder notation and a pair of parametric equations. However, it is
not true that every pair of paTametric equations represents a line. For
example, the set
S = {(*, y) : x = 2 4 k ■ 0, y = 3 + k ■ 0, k is real}
is a set whose only clement is (2, 3). Note that in the statement of The
orem 11.7, the points A and B are distinct; hence not both of the co
efficients of k in the parametric equations can be zero. That is, we can
have x 2 — xi=0 or y2 — tji=0, but we cannot have both x$— xj = Q
and y2 — yi = 0.
470 Coordinates In a Plane Chapter 1 1
Although not eveiy pair of parametric equations represents a line ,
our next theorem provides us with a method for identifying those pairs
of parametric equations that represent lines.
THEOREM 11.8 If a t h, c, d are real numbers, if b and d are not
both zero, and if
S = {(*, y) : x = n + M, y = c + dk, k is real},
then S is a line.
Proof; Taking k = and k = 1, vvc get two points in S, namely
A (a, c) and B(a + b t c + rf )• Tty Theorem 11.7, parametric equations
for AB are
sc = flr*ft(aHfc — a) = a + M
and
tfs«rx(6+ac)s@fJ&
Therefore A3 = {(x f y) ; x = a + bkj y = c \ dk,kis real};
hence S = An and S is a line.
hi addition to being able to write parametric equations for a line,
we can also write parametric equations for a segment or a ray if we
place the proper restrictions on the parameter k. We illustrate this
technique in the following examples.
Example 2 Let A = (4, 1) and B = (2, —3). Using coordinates and
parametric equations, express (1) AB, (2) AB. and (3) AB.
Solution;
1. Substituting the coordinates of A and B into the equations of
Theorem 11,7, we get x = 4  2k and y = 1  4k as para
metric equations for AB. Therefore
AB ' = {(x, y) : x = 4  2k, y = I  4k t k is real}.
2. If P is a point on AJ?, then P £ A? if and only if Jt > 0, Therefore
AB = {(*. y) : * = 4  2fe y = 1  4^ * > 0},
3. If P, A, B are three points on AB and if &i, & 2 , *a are the coordi
nates of P, A, /I, respectively, then Pis between A and # if and
only if fci is between feg and fife. Since the ^coordinates of A and
B on AB are and 1 , respectively, then P is between A and B if
and only if k is between and 1. Therefore
m = {(x t y) : x ^ 4  2k >y = I  4^0 < k<, } }.
11.6 Parametric Lln«ar Equations 471
Example 3 Given A = (4, 1} and B = (2, 3), find P on AB such
that AP = 4 'AB,
Solution: Taking k = 4 in the parametric equations of Example 2,
we get
P = fry) = (4, 15).
Example 4 Given A = (4, 1 ) and B = (2 t  3), find P on opp AB such
AP = 4 • AB,
Solution: Taking k = — 4 in the parametric equations of Example 2,
we get
P = fr y) = (12, 17).
Example 5 Given A = (4, I) and B = (2, 3), find C and D % the
points of bisection of AB.
Solution: Taking fe = j and ft = J in die parametric equations of
Example 2, we get C = (J£ , — §) and D a (, }) as the points of
bisection.
EXERCISES 1LG
In Exercises 15, the coordinates of two points A, B are given. Use para
metric equations and setbuiJdcr notation to express AB t AB, opp AH, and
m
L A = (1, 4), B = (3, 7) 4, A = (2, 3), B = (0, 0)
2. A = (2. 2). B = (5, 5) 5. A = (3, 2), 3 = (0, 1)
3. A =(1,3), B = (3,0)
GIO* Find the coordinates of the midpoint of AB in Exercises 15. {Hint:
Let k = j in the parametric equations for AB.)
1 1. Find the points of bisection of AB in Exercise 2.
hi Exercises 1217, a relation between AP and AB is given, If
A = (2, 5) and B = (4. 1), find the coordinates of P in Z8.
12* AP = 2AB 15, AP = t^2AB
13. AP = 25AB 10. AP = LSAff
14, 4AP = MB 17. AP = \AB
472 Coordinates in a Plane Chapter 1 1
1823. The instructions for Exercises 1823 arc the same as for Exercises
1217, except that F is in opp AB, Kecall that j^ = —kin this case.
In Exercise 18. AP = 2AB as in Exercise 12; in Exercise 19,
AP = 25AB as in Exercise 13; etc.
24. Find the coordinates of P £ AB if A = (1. 5), B = (4, 3), and
AP = 3A#. (There are two possible answers.)
25. Given A = (0, 4), B = (3„ 0), and (7 is on AB, find the ^coordinate of
C if its xcoordinate is — 2. (Hint: Obtain the parametric equations for
AB and let x = —2 in one of these equations (Which one?). Solve this
equation for k and use this value of k to find the (/coordinate of C.)
2a Given A = (  1, 3), B = (2, 3). and C is on A$ t
(a) find the {/coordinate of C if its xcoordinatc is 5.
(b) find the abscissa of C if its ordinate is 8.
In Exercises 2731, draw the graph of the set.
27. {(*, ij)\\+3k,y = 2 k, k is real}
28. {(x, y}:x = 3k.y = k,0<k<3}
29. {(x, y):x= 2 + k,y = 2k, k > 0}
30. {(*, y) : x = K y = K * <, 0}
31. {(x, y) : x = 5, y = 2 + Jfc, 2 < k < 3)
Exercises 3239 refer to the triangle whose vertices are A = (3, 2).
B = (9, 4), and C = (5, o),
32. Kind the coordinates of D, the midpoint of AB.
33. Find the coordinates of E, the midpoint of BC,
34. Find the coordinates of t\ the midpoint of AT.
35. Find the coordinates of the point R in CzJauch that CR s 5CD. (Use
the parametric equations for CIJ, where C =■ (xi, yi) = (5, 8).
D = (x 2 , i/ 2 ) = (6, 3), and A = f )
36. Find the coordinates of the point S in BF such that #S = JBF. (Use the
parametric equalions for BF, where B = (xt, t/i), F = (x 2 , 1/2), and
37. Find the coordinates of the point Tin AE such that AT = JAE.
38. Is Jt = S = 77
39. Show Uiat tlie three medians of A ABC intersect in a point whose dis
tance from each vertex is twothirds of the length of the median from
that vertex.
40. challenge problem. If a, b, e are positive numbers and if A = (0, 0),
B == (a, 0),. and C = (b, c), show that the three medians of A ABC are
concurrent at a point whose distance from each vertex is two thirds of
the length of the median from that vertex.
11.7 Slope 473
11.7 SLOPE
In this section we develop the idea of the slope of a line. Slope
corresponds to the idea of the steepness of an inclined plane or the
steepness of a stairway. If all the steps of a stairway are uniform, we
may describe the steepness of the stairway in terms of the "rise" and
"run** of one of its steps. The steepness of a uniform stairway is the
number obtained by dividing the rise by the run of one of the steps.
For example, we may say that the steepness of the stairway shown in
Figure 1121 is .
Figure USl
In our formal geometry, we define the steepness or slope of a line
in a similar way. That is, we define the slope of a line in terms of the
slope of any segment on the line and we define the slope of a segment
in an xyplane in terms of the coordinates of its endpointa. In informal
geometry, we can think of the slope of a segment in somewhat the
same way as we think of the slope of a step in a stairway; that is, in
terms uf its "rise divided hymn." Figure I 122 suggests that the slope
of AJ9 is $ = J. Note that in terms of the coordinates (2, 3) and (8, 7)
of the endpoints A and B of AB, the rise is 7 — 3 = 4 and the run
is 8  2 = a
Figure 1 122
kj
 1 1"
V*M
H
Ri
j*
90
I
3
. 1 .
A(*3)
Run C{8, 3)
v k
9
Coordinates in a Plane
Thus, if A = (x±, 1/1) and B
the slope m of A3 as
Chapter 11
= (x^y j/y), where *i =?^ *2» we c.otdd define
m =
x a — Kl
hut we do not. The same formula without the absolute value symbols
is not only easier to handle but it is more useful Hie sign of the slope
Indicates whether the segment "slopes up or down." In Figure 1122,
segment AB has a positive number for its slope and we note that the
segment slopes up as we view it from left to Tight. However, the seg
ment CD shown in Figure 1 123 slopes down as we view it from left to
right

'■•
111 I
ioMB
TCI
—
a
1 \
! \
BWU \
D(7 r 1)
J l
B
1  * ■
Figure 1143
If we use the formula
x 2  Xl
in compute tlie slope of CD, we obtain
15
74
4
3'
a negative number. This is one of the reasons why we do not include
the absolute value symbols in our formula for the slope of a segment.
That is, if the slope of a segment is a positive number, the segment
slopes up when viewed from left to right, and if the slope is a negative
number, the segment slopes down when viewed from left to right. We
are ready now for our formal definition.
Definition 11.2 If A(xi, y%) and B(% j/a) are two distinct
points and if Xx =£ x*, then the slope of AB is * /g "~ * 1 .
Xi — Xi
If A(xu tfi) and B{x% t tjz) are two distinct points and X\ = #2, then
"KB is a vertical segment and slope is not defined for 'KB in this case.
11.7 Siope 475
If A(*i, i/i) and B(x2, 1/2) are two distinct points and r/j = y 2 * then^B
is a horizontal segment and its slope is zero.
Note that in computing the slope of a segment A B it docs not mat
ter wiiich endpoint of AB is designated (afe 1/]) and which endpoint is
designated fa, 1/2) Thus, for the slope m of AB in Figure 1122, we
could write
m = y*_zJ± = LzJL = 1 = 1
' x 2  xi 8  2 "" 6 "" 3
or we could write
__ j/g  yi 37 _ _4 _ 2^
X2  Xi " 2  8 " 6 ~ 3 *
Similarly, for the slope m of CD in Figure 1 123, wc could write
Xz — Xx 74 3
or
Xb — xi 4—7 3
As suggested earlier, the concept of the slope of a line is based on
the concept of the slope of a segment. It seems reasonable that the
slopes of all segments of a nonvertical line are equal, and we state this
as our next theorem.
THEOREM LL9 The slopes of all segments of a nonvertical line
are equal.
Proof: Suppose that AB is any nonvertical line with A = (x u y t ) and
B = (x$, ys). Then, by Theorem 11.7, AB can be expressed parametri
cally as
= {(x, y) : x = xi + k{x 2  x T )» ij s t^  % 2  ^i), fc is r
By Definition 1 1 ,2, the slope of AB is
X 2 — XL
Let R and S be any two distinct points of AB. There are two distinct
numbers k] and h$ such that
R = (xi + fei{xa  *i), i/! + Ai(y 2  J/i))
and
476 Coordinates in a Plane Chapter 1 1
By Definition 11.5, the slope of ESh
fa + k 2 (y 2  iji))  (yi + Ky 2  j/i)) _ (</a  t/i)fe  fr)
(*"i + k 2 ix 2 — xij) — (x\ + k\(x2 — *i)) {x 2  xt)(ki  ki)
X2  «J
We have proved that any two segments of AB have the same slope, and
!hr pn lof is complete.
Definition 11.3 The slope of a oonvcrtical line is equal to
the dope of any of its segments. The slope of a non vertical
rav is equal to the slope of the line that contains the ray.
Example 1 Find the slope of the line which contains the points
A = (2,5) and B = (4,0).
Solution: AB is a nonvertieaJ line and, by Definition 11.3, the slope
of AB is equal to the slope of A~B. Therefore the slope of AH is
y 2  *Ji _ 05 _ _£
*2*i "4<2)~ 6'
We know that two distinct points determine a line in the sense that
if any two distinct points are given, then there is exactly one line which
contains them. It is also true that a non vertical line is determined by
any point on it and its slope. That is, there is exactly one line which
contains a given point and has a given slope. We now state tins
formally,
THEOREM 11.10 Given a point A and a real number m » there is
one and only one line which contains A and has slope m.
Existence. Let a point A = (*i, yi) and a slope m be given. Let
B be the point (x\ + 1, «/i 4 m). Then AH is a line which contains A
and has slope m. (Show that the slope of AB is m.)
Uniqueness. Let PQ be any line which contains A and has slope
m. Since FQ is a nonvertical line (Why?), it intersects the vertical line
through B(xi + I, y 1 f m) in some point R(xi — 1, y 2 ) as suggested
in Figure 1124. (The figure shows R and B as distinct points. We shall
prove that they are actually the same point.) The slope of FQ is equal to
11.7 Slope 477
Figure US4
Therefore y 2 = jft + «b Hence H = B and Py sa AB, This proves
that A/5 is the only line through A with slope m, and the proof is
complete,
If wc know the coordinates of two points on a line, we can use The
orem 11.7 to write parametric equations for the line. Theorem 11.10
implies that a line is determined by any point on it and its slope. There
fore we should be able to write parametric equations for a line passing
through a given point and having a given slope. Our next theorem tells
us how to do this.
THEOREM I hi I The line I given by
1. I s {(*, y) : x = *i + fc, y = y, f mk, k is real}
or by
2. I = {{x, y) : x = xi + sk t y = y t + rfc, k is real)
is the line through (x lt y\) with slope m = — .
Fnx>f: We shall establish (J) and (2) separately.
Troof of 1: Taking k = and k = I in the parametric equations in
(1), we get (x% t i/i) and (*j f 1, y± + m), two points on I. The slope of
the segment joining these two points is
yi + m  yi _rn_ m
*i + 1  xi 1
Therefore I as given in (1) is the line through (%, i/i) with slope m.
478 Coordinates in a Plane Chapter 11
Proof of 2: Taking k = and k a i in the parametric equations in
(2), we get (*i, yi} and (x t + $, y^ + r), two points on I The slope of
the segment joining these two points is
j/i + r  yi _ r
*i + &  Xy $
Therefore I as given in (2) is the line through {x^ t/i) with slope rn, and
the proof is complete.
Example 2 A line p passes through (2, 5) and has slope 3. Find the
point on p whose abscissa is — 2.
Solutimt: Using Theorem 11,11, we can express p as
p = {(x, y) : x = 2 + ft, y = 5 + 3fc, k is real).
We next set % = 2, obtaining 2 = 2&,orfc= 4. Hence
y = 5 + 3(4) = 7.
The point on p whose abscissa is —2 is (— 2, —7).
Example 3 Find the slope of the line
I = {(x, y) : x = 2  3fc # = 3, Jfc is real}.
SoJi*rio»: Taking k = and k = 1, wc get (2, 3) and ( — 1, 3), two
points on the line. The slope of the segment joining these two points is
3  3 _
12 3
Therefore the slope of f is zero. Is f a vertical, horizontal, or oblique
line?
We conclude this section with the following summary remarks re
garding the slope of a line.
L If a line has a positive number for its slope, then the line is ob
lique and slopes up when viewed from left to right, (See Figure
ll25a.J
2. If a line has a negative number for its slope,, then the line is ob
lique, and slopes down when viewed from left to right, (See Fig
ure ll25b.)
3. If a line has slope aero, then the line is horizontal. (See Figure
ll25c.)
4. Slope is not defined for a vertical line.
11.7 Slope 479
Zero
■lOM
(C)
Figure 11
EXKRCI5ES 11.7
In Exercises 110, find the slope of the segment joining the given >oints.
Express each answer as a fraction in lowest terms.
1. (2, 5) and (5, 7) & (2^ 4) and (1, 4\)
2. { 1. 3) and (8, 3) 7. (5.6, 3) and (1.4, £)
3. (0, 4) and (8, 0) 8. {1.2. 5) and (3.2, 5)
4. ( 1, 4) and (2, 2) 9. (2, 5) and (5, 2)
5. (3, 3) and (3, 3) 10. (7. 12) and (1, 3)
11^20. In Exercises 1120, tell without plotting the points whether the
given line (a) slopes up, (b) slopes down, or (c) is horizontal. In Kxercisc
1 1, the line is the one that passes through the two given points in Ex
ercise 1. In Exercise 12, die line is the one that passes through the two
given points in Exercise 2, and so on.
21. (a) On the same xt/plane. graph the line r through the two points of
Exercise 1 and the line s through the two points of Exercise 2,
(b) What appears to be true about these two lines, that is, how are r
and ^ related?
(c) How are the slopes of these two lines related?
22. (a) On the same xyplane, graph the line p through the two points of
Exercise 3 and the line q through the two points of Exercise 4.
(b) What appears to be true about these two lines, that is, how are p
and q related?
(c) How arc the slopes of these two lines related? [Hint: Find the prod
uct of the slope of p and the slope of q.)
In Exercises 2327, find x or y (whichever is not given) so that the line
through the two points will have the given slope,
23. (4, 1) and (x, 4), m = 3 26. (4, 3) and (0, y), m =s 
24. (3, 1) and (6 t y) t m s \ 27. (5, 12) and ( 2, y), m 
25. (x, 0) and (3, 5), m = J
480
Coordinates in a Plana
Chapter 11
28. (ti) On the same xtfplane. graph the line I through the two points of
Exercise 25 and the line n through the two points of Exercise 2ft
(b) What appears to be true about these two lines, that is, how are J and
n related?
(c) How are the slopes of I aud n related?
29. (a) Plot the quadrilateral ABCD with vertices A = (— 2, —3),
B = (3, 2), C = (4, 2), and D = (~1. 1).
(b) Which pairs of sides have the same slope?
(c) Is A BCD a parallelogram?
30. Write parametric equations for the line I through { — 2, 5) with slope 2.
31. Write parametric equations for the line p through (1,  1) with slope J.
32. Use the parametric equations for line / obtained in Exercise 30 to find
the coordinates of at least one more point on I and plot the graph of 1.
33. Use the parametric equations for the line p obtained in Exercise 31 to
find the coordinates of at least one more point on p and plot the graph
of p.
34. One way to plot the graph of line p in Exercise 31 is suggested by the
figure. Note thai the point Q{1,  1) is on p and the "slope fraction" f
tells us how to get from one point to another on p. The numerator 3
represents the difference in the ordinates of two points on p, and the de
nominator 2 represents the difference in the abscissas of the same two
points. Thus, if we begin at the point (1, —1). which is known to be
on p, we arrive at another point on p by increasing the ordinate and
abscissa of [L — 1) by 3 and 2, respectively. If the slope is negative, it
can be written as a fraction with a negative numerator aud a positive
denominator; for example, —^—, In tliis example we could get from one
point to another, in informal geometry, by moving 2 to the right and 3
down; in formal geometry, by adding 2 to the abscissa and subtracting
3 from the ordinate. Write the coordinates of the points A T B T C.
.
":•
fp
% /
/
: ; i
s
2
a
1
/
J
\
a
/ x %
m
■
1)
/I
11.8 Other Equations of Lint* 48 1
35* Use the method described m Exercise 34 to plot the graph of the line
through (—2, 1) with slope y.
36, Use the method described in Exercise 34 to plot the graph of the line
through (0, 6) with slope J,
37, Use the method described in Exercise 34 to plot the graph of the line
through (0. 0) with slope 2. (ffinf; 2 = f )
11.8 OTHER EQUATIONS OF LINES
In Section 11.6 we showed how a line can be expressed in terms of
parametric equations. In this section we shall show how parametric
equations for a line can lie used to obtain other equations of the line.
You are already familial" with most of these equations from your work
in algebra. Perhaps the simplest equations for lines are those for hori
zontal and vertical lines.
THEOREM 1L12 If Ms the horizontal line through {x lf 1/1), then
'= tfey) '■ y = »)■
Proof: Let (jcfc, 1/2) he any other point on the horizontal line I through
the point [t\, iji). By Theorem 11.7,
1 = (fr y) : * = *i + H*2  *i), y = yi + %s  yi)> k k real}.
Tine I is horizontal so tj\ = y 2  Therefore
y = yi + %2  #1) = !/i + *0 = y T .
For any real number fc»
* s x* + k(x2 — x%)
is a real number. Conversely, for every real number ,t» there is a real
number k such that x = Xt + fc(*2 — x i) Th^ one an ^ on ly ^ that
works here is
*2  ^l
It follows that I is the set of off points (at, «/) such that j/ = r/ t and x is a
real number, that is, that
This completes the proof.
462 Coordinates In a Plane Chapter 1 1
Iii the statement of Theorem 11.12 it should be clear that we could
say that
1 = {(*> y)  x m Xi + k{x 2  Xi), y = i fl , k is real},
or
I = {(x t if) : x is real and ij = yi} t
or
' = {(*; ST : y = §Ji. * is real}*
Since the equation x = xi + A(* a — *i) establishes a onctcotie cor
respondence between the set of all real numbers thought of as k values
and the set of all real numbers thought of as xv&lucs. these setbuilder
symbols all denote the same set.
THEORTM 1L13 If I is the vertical line through {x u ft), then
/ = {(x, y) ; x *i}.
Proof: Assigned as an exercise.
Example 1 If P = { — 4, 3), write an equation of (1} the vertical line
through P and (2) the horizontal line through P.
Solution*
1. An equation of the vertical line through Pis x = —A.
2, An equation of the horizontal line through P is y = 3.
Our next theorem tells us how to write an equation of a line if we
know the coordinates of any two distinct points on the line.
THEOREM 11.14 {The TwthPoint Form) Tf A = (x u y\) and
B = (x2, 1J2) are distinct points, and if AB is an oblique line, then
= {fr„) :£=*= JLzIl)
I X 2 — Xi 1/2  Ui)
*2  x t y 2  y^
Proof: We are given that A = {xi, tji) and B = (x z , ife) are distinct
points on oblique line AB< There are two things to prove.
1. If P = (x, u) is a point of Aft, then — = " "~  1  .
2  If ^3^ = ^7=^ ^ P = ^ $ '^point of £2.
UJB Oth« Equations of Lin** 433
Proof of 1; Suppose that F a (x, 1/) and that P is a point on A/3, If
P = A, then
JL^L = . and JLZliL^XlllL.
if*  yi *2  *i y*  yi
If P ^ A, then by Theorem 1 1 .9 the slope of AT equals the slope of AS
since AP and A B are segments (not necessarily distinct) of the same line
aS. But the slope of £B is ^ ~ * jl and the slope of A? is ^~ & .
x 2 — Xi x — Xj
Therefore
ya¥i _ gffi
X 2  X, "" X  Xi "
Multiplying both sides of this last equation by x — Xi, then dividing
both sides by y 2 — y& then simplifying, we get
*  *i _ y  yi
*2 — %i 1/2  yi "
Therefore, in all cases, whether A = P or A=?=P t 'd P £ AB, then
*  *i _ y  tfi
ac 2  xi y a  1/1
Proof of 2: Suppose that P = (x, 1/} and that
xxi _ y yi
Xi  x, " y 2  yi "
Either * — Xi = or x — x± ^= 0. If x — xi = 0, then 1/ — j/ a = 0,
(a, y) = (xi, yi), P = A, and P € AB, If x — x, ^£ 0, multiply both sides
of
**t _ yyi b y 2  yi
*2  xi Jfa — yi x  xi
to get
Uzlfi yUi
Xg — Xi X — Xl
Since ~ 5EL is the slope of AB and *• *± is the sloi>e of AP, this
#2 — *l X — Xi
proves that if x — x x =j£ 0, then the slopes of AB and AT 5 arc the same
and, as we shall prove, P Q AB.
484 Coordinates in a Plane
Chapter 11
Suppose, contrary to what wc assert, that the slopes of A H u i id . \ / '
are equal, but that A } B, P are noncollinear as in Figure I. .1.26.
Figure 1126
Let Qfe y') be the point in which the vertical line through P
intersects AB. It follows from Theorem 11.9 that the slopes of A%) and
AB are equal, so the slope of A$ is equal to the slope of AP, and
y* y\ _ y yi
X — Xt X — JTi *
y'  ui = y  y^
ft = y>
and the points A, B, P are collinear. Since our supposition that A, B, P
are noncollinear leads to a contradiction, this proves that our suppo
sition is false. Therefore A, S, P are collinear and P lies on AB. This
completes the proof in all cases, whether x — Xi =: or x — X\ y^ 0,
that if
*  *i _ y  yi
*2  *i "" y*  yi
and P = fc y) t then F £ AB. This completes the proof of Theorem
11.14.
The equation
*  *i _ y
yi
yi yi
in the statement of Theorem 1 1 . 14 is often written in the form
(i) y  yi = *±=3Ht  x,).
*2 — 3.1
11.8 Other Equations of Lines 485
Show how Equation (1) is obtained from the one in the statement of
Theorem 11.14. Since A = (acj, y\) and fi = {x^ y%l are distinct points
on the oblique tine AB of Theorem 11.14, then
2i = m
— *1
is the slope of AB. Substituting m for ^ ^ in Equation (1), we
(2) rj  y t = m(x  xx)
for AB. It should be noted that Equation (2) still holds if XS is a hori
zontal line since, in this case, m = and Equation (2) reduces to
y = yi. Thus, if we know the slope of a line and the coordinates of any
point on the line, we can write an equation of the line. When an equa
tion of a line is written in the form of Equation (2), it is often referred
to as the point slope form. We have proved the following theorem,
THEOREM 1L15 (The Point Slope Form) If Hs the line through
A = (xu yi) with slope m, then
'={(*> y) ' v  *j\  »*(*  *i)}
Example 1 If A a ( — 2, 3) and B = (4, ft), write an equation of
Solution: Substituting the coordinates of A and B in the TwoPoint
Form of an equation, we obtain
*  (2) y3
4  (2) 63
or
x 4 L 2 _ y 3
6 ~ 3
as an equation of AB.
IF we multiply both sides of the equation in Example 1 by 6, and
add — 2y 4 6 to both sides, we obtain
x  2y + 8 = t
which is of the form
Ax + By + C = 0,
with A = 1, B = 2, and C = 8.
486 Coordinates "in a Plane Chaptarll
This latter form is often referred to as the general form of a linear equa
tion, that is, of an equation whose graph is a line. Although we shall not
do so here, it can be proved using the theorems of this chapter that if
A, £, C are real numbers with A and B not both zero, then Ax + By +
C = Ois an equation of a line. The converse statement is also true, that
is, every line has an equation of the form Ar + By + C = in which
A, B, C are real and A and B arc not both zero. What is the graph of
Ax + By + C = if A a 0, B = 0, C=£0?ifA = 0\ B = 0\C = 0?
Example 2 Write an equation of the line through (3, —4) with slope
— ■§ and put the equation in general form.
Solution: Substituting (3, — 4) for {%%, grj and —  for m in the Point
Slope Form of an equation, we obtain
i, + 4 = <r  3).
Multiplying both sides of this last equation by 3 and adding 2* — 6 to
both sides, we obtain 2i + 3i/46 = 0asan equation of the line in
general linear form.
Every nonvertfcal line intersects the ysoas (Why?) in a point
whose coordinates are (0, h), where b is a real number. The number b
is often called the y intercept of the line. Similarly, every nonhorizontal
line intersects the zaxis in a point whose coordinates are (a, 0), where
a is a real number. The number a is called the xmtercept of the line. It
should be clear that die x and {/intercepts of a line can be obtained
from an equation of the line as follows:
1. If the line is a nonhorizontal line, the ^intercept is obtained by
substituting for tj in an equation of the line and solving the
resulting equation for x,
2. If the line is a tionverrical line, the (/intercept is obtained by
substituting for x in an equation of the line and solving the
resulting equation for y.
Example 3 Find the x and t/tntcrccpts of the line whose equation is
3s  4y + 8 = 0.
Solution: Substituting for y i u the equation 3x — Ay f 8 = 0, wc
obtain 3x + 8 = 0, or x = — 2$. Hence the ^intercept is — 2. Sub
stituting for x in the equation, we obtain — Ay + 8 = 0, or y = 2.
Hence the {/intercept is 2.
If we substitute (0 T b) for {%i, ij\
equation of a line, we obtain
in
11.8 Other Equations of Unas 487
the Point Slope Form of an
or
y — b = m(x — 0)
tj = mx + b.
It should be clear that when an equation of a nonvertical line is put in
the form ;/ = mx + b, the slope and (/intercept of the line can be read
directly from the equation, This form, which is often called the slope
yintercepl form, is especially convenient when one wants to draw T tlie
graph of a line whose equation is given.
Exanipte 4 An equation of the line / is 3* + 2y — 8 = 0.
L Put the equation in slope (/intercept form.
2. Draw the graph of I on an xyp\m\c.
Solution:
1. To put the equation 3x + % — 8 = in slope (/intercept form
we add — 3x 4 8 to both sides and then divide both sides by 2.
The resulting equation is u = — Jar + 4.
2. The graph of t is shown in Figure 1 127. Note that the slope and
(/intercept of I can be read directly from the equation y —
— Jx ( 4, Thus, if we start at the point where y = 4 on the
yaxis [that is, the point (0, 4}] and "trace out" a slope of — §,
we arrive at a second point on /, These two points determine
/ and henee they determine the graph of I.
I

V>
».4)1
5 2
C~ i
 3
V
\ X
5
JV_ 5
Figure 1127
As you might expect, two nonvertical lines are parallel to each other
if and only if they have the same slope. If you worked Exercises 22 and
28 of Section 11.7, you should have discovered a relationship between
tlie slopes of two lines that are oblique and perpendicular. We state
these properties of parallel nonvertical lines and of perpendicular
oblique lines as the last two theorems of this section.
488 Coordinates in a Plane
Chapter 11
THEORESf 11.16 Two non vertical lines are parallel if and only
if their slopes are equal.
Proof; Let two non vertical lines r and 9 be given. We have two things
to prove.
L If r  $, then their slopes are equal.
2. If the slopes off and s arc equal, then r s.
Proof of 1: Suppose that r and * are nonvertieal lines and that r [f $, If
r = s T then r and s are the same line and hence they have the same
dope. Suppose r^=s t as shown in Figure J J 28. Let l\xu yi) and
Q(xt, 1/2) be two distinct points on r. Let the vertical lines through
P and Q intersect s in F[xi, iji + k) and Q'{x 2> j/2 + k), respectively.
Figure 112H
Then PFQ'Q is a parallelogram (Why?) and PF = Q^. But PF = \h\
and QQ 1 = k\. Why? Therefore \h\ = \k\. Since h and h are either both
positive or both negative, wc have h = L The slope of Pfp (and hence
of r) is
*2 — *i
.The slope of FQ' (and hence of s) is
(jfe + A)  (yi 4 h) _ y 2 y 1
%2 — *! X2  Xj '
since ft = fe Therefore the slopes of r and s are equal.
Proo/ o/2: Suppose that r and s are nonvertieal lines and that their
slopes are equal. If r = s ( then by definition r I .9. Suppose that r =£ s.
Let m be the slope of r. Then s also has slope m. Now, either r and s arc
11.8 Other Equations of Lines 489
parallel Of they are not. Suppose they are not parallel. Then they have
exactly one point P(a"i, r/i) in common. Tin is we have two distinct lines
passing through the same point and having the same slope. This con
tradicts Theorem 11.10, Therefore our supposition that r and s are not
parallel is incorrect; hence it follows that r  $ and the proof is
complete.
THEOREM Il.il Two oblique lines are perpendicular if and
only if the product of their slopes Is — 1.
Proof. Let lj and Z 2 be two given oblique lines with slopes mi and wig,
respectively. We have two statements to prove.
1. If /i _L h, then mi • mg = — 1 .
2, Umi*m 2 = — 1 , then h _ k
Before proceeding with the proof we wish to comment on the
adjective "oblique" in the statement of the theorem. Would the state
ment that results if "oblique" is erased be a theorem? No, it would not.
For if one of two lines is not oblique, then those two lines arc perpen
dicular if and only if one of them is horizontal and the other is vertical.
Since a vertical line has no slope, there would be no product of slopes
in this case.
We now proceed to the proof of statements 1 and 2.
Let p\ and p% be the lines through (0, 0) and parallel to h and fa,
respectively, as shown in Figure 1129. Then the slope of pj is mi and
tie slope of p% is nio. Why? Since neither »i nor »2 is a vertical line,
they both intersect the line
I = (fc tj) : x = 1},
Figure 1 123 p\
490 Coordinates in a Plane
Chapter 11
Let A = (1, gft) and B = (1, 1/2) be the points of intersection of I with
lines pi and p 2 , respectively, as shown in Figure ] 130,
Figure 1130
Tli en the slope of p\ is
and the slope of p 2 is
Wi 0
y 2 0
m 2 = *— ~ = y 2 .
Therefore A = (1, mi) and B = (I, ma). Now, h _L l 2 if and only if
pi 1 pz
It follows from the Pythagorean Theorem and its converse (applied
to AOAB in Figure 1130) that
Pl _ p 2 if and only if (OAf + (OB)* = (AB) 2 .
From the distance formula, we get
(QA)* = 1 4 wn 2 ,
(OB) 2 s 1 + mtf*
and
Thus
(AB) a = (ms  mi)*.
if and only if 1 + mi 2 ■+■ 1. 4 m 2 2 ss (mg — mi) 3 ,
if and only if 2 j wi 8 + «»^ = m 2 2 — 2m im^ + »»i a ,
if and only if 2 = — 2f9t3!fft&
if and only if mim? = — 1,
Therefore £1 _L Z 2 if and only if m jm 2 = — 1, and the proof is complete.
11.8 Other Equations of Lines 491
Example 5 If
h =((*,!,): 5* 2y + 4=0},
h = ((jtv y) : 2a: + 5y  15 = 0},
and
fe ■ {(a& </) : 5*  2#  8 = 0},
show that l\ I ?3 and that ti ± h.
Solution; Putting the equations of I it f 2 , /g in slope ^intercept form,
we get
h  {{*> V) : </ = $* + 2}»
I 2 = {(*,</): y = — $* + 3}>
and
k = {(J, y) : y = fx  4}.
Since ^ = slope of li = slope of h> it follows that 1$ \\ I3, Also, since the
product of the slopes of l\ and 1% equals
*•(« = i.
It follows that /i X fe, K »s also true that (3 J k Why?
EXERCISES 11.8
1. Prove Theorem 11.13.
In Exercises 26, use the TwoPoint Form to write an equation of the line
containing the given points, and put each equation in general form (that is,
the form Ax + By + C = 0).
2. (1, 5) and (3, 4) 5, (<X 0) and (  1, 6)
3. (0. 3) and (5, 0) 6. (2, 2) and (2, 2)
4. (0 t 3) and (5, 0)
In Exorcises 712, use the Point Slope Form to write an equation of the line
which contains the given point and has the given slope, and put each equa
tion in general form.
7. (3, 5) and m = 1
8. (2, l)andm= 1
9. (a0)andm = 
10. (5, 0) and m = 
11. (3, 7)audm = 1
12. (5, 3)andm = 
492 Coordinates in a Plane Chapter 1 1
■ In Exercises 1316, determine which word, parallel or perpendicular, would
make a true statement.
13. The lilies of Exercises 7 and 8 are T] .
14 The lines of Exercises 8 and 1 1 lire [Tj and distinct.
15, The lines of Exercises 9 and 12 are [TJ . Are these two lines distinct?
16. The lines of Exercises 9 and 10 are [?]
17* Write an equation (in general form} of the line which contains the point
{3, 8) and is parallel to the line whose equation is 2x — 3y = 10. (Hint:
What is the slope of the line whose equation is 2* — 3y = 10?)
18. Write an equation (in general form) of the line which contains the point
;3, Si and is perpendicular to the line whose equation is 2x — Zlj = 10.
19. Write an equation (in general form) of the line which contains the origin
and is perpendicular to the line whose equation is y = x.
■ Iu Exercises 2025, an equation of a line is given. In each exercise, (a) put
the equation in slope (/intercept form, (b) find the slope of the* line, and
(c) find the x and ^intercepts of the line.
20. 2x  3y  12 =
21. 3x + 2y = 16
2& 4x  6y = 8
23* y + x =
24. 4x  2ij = 1 1
25. 5x + 4y + 13 m
26. Which pairs, if any, of the lines of Exercises 2025 are parallel? Which
pairs, if any are perpendicular?
27. Find, without graphing, the coordinates of the point of intersection of
the lines
P = {{*> a) : * + % = 6}
and
q={(x,y):5x + 4tf= 3}.
(Hint: Note that p is not parallel to q (show this) and hence the lines
intersect in exactly one point.) l>et (x u y±) be the point of intersection
of lines p and q. Then, since (*j, j/j) is on p, x x + 3i/i = 6 or
(1) Ui m  j*i + 2.
Also since (x\, y^) is on q ¥ Sxj f 4yi = —3 or
m sn = ^  1
Apply the substitution property of equality to Equations ( 1} and (2) and
find xi and tji.
1U8 Othar Equations of Lines 493
28. Show without graphing that lines
p={(x > t,):2x3y=i2}
and
q = fc y)  4y + x m 5}
are not parallel, and find the coordinates of their point of intersection,
(See fjcercise 27.)
Exercises 2939 refer to the triangle whose vertices are A = (1, 1),
B = {9, 3), and C = (7 t 9). In each case where an equation of a line is asked
for, write the equation in slope (/•intercept form. (It may help to draw the
figure in an xyplane and label the points as you need them.)
29. Find the coordinates of the midpoint D of AB.
30. Find the coordinates of the midpoint £ of BC.
31. Find the coordinates of the midpoint F of AC*
32. Write an equation of the line through A and &
33. Write an equation of the line through B and F.
34. Write an equation of the line through C and D,
35. Show that the lines A£. Bb\ CD intersect in the same point.
36. Write an equation of the line through E and F.
37. Write an equation of the line through A and B.
38. Show that F?  XS and hence that bW  AB.
39. Show that EF = \AB.
40. If a # and h s£ 0\ an equation of the form J f = 1 is called the
ah
interrupt form of an equation of a line. Show that the line whose equa
X ft
tion is r i — l contains the points (a, 0) and (0, b).
41. Put the equation 3* f Ay = 12 in intercept form (see Exercise 40) and
read the .v and [/intercepts directly from the equation.
42. Given that p = ((x t y) : x = $},
(a) Is(3 5 7)€p?
(b)ls(3,l7)ep?
(c) b(4,4)Cp?
(d) Is (3, 9) € p?
43. Given that q = {(a, y) : y =  i}:
(a) Is(4,5)€<j?
(b) Is(4, V5)€<??
(c) Ik(4,w)€tf
(d)Is(4,4)Cc/?
494
Coordinates in a Plane
Chapter 1 1
44. challenge problem. Use theorems of this chapter to prove the fol
lowing statement: If A. B, C arc real numbers with A and H not both
zeru T then Ax + By + C = is an equation of a line.
45. challenge problem. Use theorems of this chapter to prove the fol
lowing statement: Every line has an equation of the form Ax + By +
C = in which A, B,C are real and A and B are not both zero. (Note
that this statement is the converse of the statement in Exercise 44.)
11.9 PROOFS USING COORDINATES
We have defined an xycoordJnate system in a plane and have used
coordinates as tools in much of our work in tills chapter, Given a plane,
there are many xy coordinate systems in that plane. In constructing a
proof of a geometric theorem, it is wise to select a convenient ^coor
dinate system that fits the problem and, at the same time, reduces the
number of symbols needed in the proof. Such a selection yields no loss
of generality, yet reduces the amount and difficulty of work involved.
We illustrate with our next theorem which appeared as a corollary in
Chapter 10.
THEOREM 11.18 A segment which joins the midpoints of two
sides of a triangle is parallel to the third side and has half the length
of the third side.
We shall give two proofs of Theorem 11.18. In the first proof,
we select an arbitrary x ^coordinate system in the plane of the triangle
without any regard to the position of the vertices and sides of the given
triangle. In our second proof, wc "pick" an ^coordinate system in
the plane of the Lriangle in such a way as to reduce the number of sym
bols needed in the proof,
Proof I: Let AABC in any aryplane be given. (See Figure 1131.)
Cfaun)
figure 1131
£(*S>«)
11,9 Proofs (Mill Coordinates 495
Suppose that
A = (xi, yi) t B = (x 2j jfe), C = (x 3} y s ).
Let D, E, F be the midpoints of BC y AC, KB, respectively. Then
U.sinu; llic Distance Formula, we get
(BE) 2 = (^_±ii _ *i ±j» V + / yg + ys _ g/i + ya \ 2 .
= ft*  *i)) 2 + (Ksfc  ^)j*,
But, by the same formula,
{AB)* = {x 2  x v f + (y,  m f.
Therefore
{DEf = i(AB)2 t
° r DE = #AB).
Next* we prove that DE [ KB. Suppose that AB is a vertical segment
Then,
xj a 3C2 Rod —  — SB — i
Therefore
"~l 2 ' 2~~ /*
F _ (*i + x 3 yi + y 3 \
and DE is a vertical segment. Hence DE  AB. If ZB is not u vertical
segment, then its slope is ^ 2 ~ ^ 1 . The slope of TM is
xs — Xi
2 2 _ y 2  yi
Xi + x s xi + x 3 x 2  xi '
Therefore, since DE and AB have the same slope, DE [ KB. This com
pletes the proof so far as the segment DE is concerned. In a similar
way, wc can_prove that EF = \BC and EF  BC and that DF = ^AC
and DF II AC.
496
Coordinates in a Plane
Chapter 11
Proof H: Let A A B C be given . 1 n the plane of this t dangle there is an
^/coordinate system with the origin at A 9 with Afi as the xaxis, with
the ^coordinate of B positive, and with the ^coordinate of C positive,
(See Figure 1132.)
q2x&2>a)
Figure 1132
B{2xi,0)
Let x lt x 2s y2 he real numbers such that B = (2x [3 0), C — (2x2, 2f/2}*
Let D t £, f* be the midpoints of EC, AC, AB, respectively. Then
D = (xi + x s , i/ 2 ), E = (x 2s y 2 )»
the slope of DE = 0, the slope of KB = 0,
DE a {X! + x 2 )  x 2  = jxj = x lt
and
AJ? = 12x41 =2*.
Therefore DE  KB and £»E a {A#. This completes the proof for UE
and this is all that we need to prove, since DE might be any one of the
three segments which joins the midpoints of two sides of the given
triangle.
To prove the statement of the theorem for the segment which joins
the midpoints of two sides, we first label the triangle so that AC and CB
are those two sides and then proceed as above. Thus each of the three
parts of the proof uses a different coordinate system, but what we write
in each case is the same. For example,. Figure 1133 shows another pic
ture of the triangle shown in Figure 1132. However, it shows a differ
ent xycoordinate system and a different labeling of the vertices. Vertex
C in Figure 1132 becomes vertex A in Figure 1 133, A becomes B t and
B becomes C, and we have made the xaxis look horizontal.
You should note that the proof that DE = \AB and DE \\ AB would
proceed exactly as before, but UR in Figure 1133 is not the same seg
ment DE as in Figure 1132. The same applies for the segment AB,
11.9 Proofs Using Coordinates **?
C(2i 2 ,2>'2)
Figure 1133
It is clear Hi at Proof II is simpler than Proof I and is to be preferred.
In general, if a proof using coordinates involves a polygon, it is usually
easier to construct a proof if we select an ^coordinate system in the
plane of the polygon satisfying one or both of the following conditions:
1 . Let the origin be a vertex of the polygon and let the positive part
of the xaxis contain one of the sides of the polygon,
2. If the polygon contains a right angle as one of its angles, let the
origin be the vertex of the right angle and let the positive parts
of the x and (/axes contain the sides of the right angle.
THEOREM 11,19 The medians of a triangle are concurrent in a
point (eentroid) which is twothirds of the distance from each ver
tex to the midpoint of the opposite side.
Proof: Let A ABC he given. Select an ^coordinate system in the
plane of this triangle with the origin at A, with AB as the xaxis, with
the abscissa of B positive, and with the ordinate of C positive. (See
Figure 1134.) Let a, h, c be numbers such that B = [Ga, 0), C =
{&b f 6c), Let D s E, F be the midpoints of BC, CA, X73, respectively.
i
* cm*;)
E(3b>3e/ J \Dr3o+36,3e>
/ Nt'" A
/uo.0}
i\3c.U) 0(60,0)
Figur# U34
498 Coordinates in a Plane Chapter 11
Then AD, BE, CFare the medians of A ABC. We must prove that AD,
BE, CF are concurrent at some point P and that AF = JAD, BP m
BE, and CP = CF.
The midpoint of AB is F(3a, 0). We can express CF parametrically
as follows:
QP= {fcy) : x = 66 + {3a  66)*, y = Gc + (0 6c)*, < * < 1}.
The point P on CF such that CP = CF can be obtained by setting
k =  in the parametric equations for CF. Thus
x = bh + (3a  66} • J
 ($6 + 2a  46
= 2b + 2a
and
Therefore
y = 6c + (0  6er) ■ §
= 6c4c
= 2c.
P= (2a + 26, 2c).
Similarly, the midpoint of J5C is
D = (3a + 36, 3e)
and
AD = {{s, • i = 4. (3c + 36  0)fc,
i/ = + (3c 0)*,0< fc < 1}.
The point F on A D such that AF = %AD is obtained by setting k = J .
Tims
s= (3a + 3/?)* 
= 2« + 26
and
f (*)*}
= 2c.
Therefore
F = (2a + 2b, 2c).
The midpoint of CA h E = (36. 3c) and
BE = (fey) : x s 6a 4 (3fr  fia)*. y = + (3c  0)*,0 < k < 1},
The point F' on BE such that BF' = §BE is obtained by setting k = § .
11.9 Proofs Ustnf Coordinates 499
Thus
and
Therefore
x = 6a + (36  6fi) ■ 
= 6a + 2fo  4c
= 2a + lb
= 2c.
P" = (2a + 2fc, 2c).
Wc have shown that
P = r = P" = (2a + 26, 2c),
that CP = fCF, AP = \AD t and that BP = %BE. Therefore the medi
ans of a triangle are concurrent at a point which is two thirds of the
distance from each vertex to the midpoint of the opposite side, and trie
proof is complete,
In the second sentence of this proof we could have taken a t h t c as
real numbers such that B = (a, 0) and C = {b t c). The resulting ex
pressions for the coordinates of D, E T l\ and P would have involved
many fractions. We avoided these fractions by taking a, 6, G so that
B = (0a, 0) and C = (65, 6c).
You may feel that Theorem 11,19 would be easier to prove without
using coordinates. It is possible to construct such a proof using the the
orems, postulates, and definitions that we have established before this
chapter. You might be interested in trying to do so.
As Indicated in the statement of the theorem the point of intersec
tion of the medians of a triangle is its ceutroid. In informal geometry
we think of it as the balance point; it is the point where a cardboard
triangular region of uniform thickness balances. In calculus the idea
of moments of mass (extending the idea of weight times distance in
teetertotter exercises) is introduced and extended to develop a theory
of centroids for plane figures. The centroid of a triangle is an example
of a centroid as the concept is developed formally in calculus.
Our last example of this section is a theorem that you will find help
ful in working some of the exercises at the end of the section.
THEOREM 11.20 Let quadrilateral ABCD ivith A = (0, 0),
B = (a, 0), D = (h> c) lie given. ABCD is a parallelogram if and
only if C = (a + b> c).
MM
Coordinates in a Plane
Chapter 11
Figure 1135 shows one possible orientation of Lhe given quadri
lateral ABCD in an xt/plaiic. However, our proof depends only on A
being at the origin and 8 being on the itaxis as given in the theorem.
We must prove two tilings.
h JfC = [a + b, c), then ABCD is a parallelogram.
2. If ABCD is a parallelogram, then C = (a 4 b, c).
D(b t e)
a**y)
^m
5:q.,u;
Figure 1135
Proof of h If C = (a + b, c), the slope of CD is
c — c _ 0. = o
a + b — b a
Also, the slope of AB = ^ = 0, Therefore CD \\ AB. Since KB and UD
ai"c horizontal segments (Why?}, we have
AB = \a\
and
CD = \a + b  b\ = a\.
Therefore AB — CD and ABCD is a parallelogram.
Proof of 2: If ABCD is a parallelogram, we must prove that
C = (a + b r c). Since ABCD is a parallelogram, KB  t25. The slope
of A J? is 0j therefore the slope of CD is 0. Let C = (x, y); then the slope
of CD Ls
= 0.
a; —
ITierefore yc = 0andy= c. We have AB = CD (Why?) and
AB = d and CD = \x  b\.
Therefore
\a\ = k  R
11.9 Proofs Using Coord f nates 501
Now, if a > 0, then x > b % and if a < 0, then * < b. (If a > and
x < fc, or o < and x > fc t then C and B would be on opposite sides
of AD and ABCD would not be a parallelogram.) It follows that
a = x — b or that ac = a ■+■ fc. Therefore
C — (a + h, c)
and the proof is complete.
EXERCISES 11.9
Unless stated otherwise, use coordinates to prove the theorems in this set of
exercises* Many of these theorems have appeared as theorems or exercises
earlier in the texL We include them here since they can be proved easily
using coordinates.
1. Prove:
THEOREM 11.21 If the diagonals of a quadrilateral bisect each other,
then the quadrilateral is a parallelogram,
(Hint: Let A = (0, 0). B = (o, 0}, C = (x, y), and D = {h, c) be die vertices
of the quadrilateral and suppose die diagonals of the quadrilateral bisect
each other. Show that x = a + b, y = a, and then apply Theorem 11,20.)
2. Prove:
THEOREM 11.22 The diagonals of a parallelogram bisect each other.
(This is the converse of Theorem 1 1.21. By Theorem 11 20, the vertices of a
parallelogram may be taken as A = (0, 0). B = (a t 0), C = (a + h, c), and
D = (h, c)+ Show that the midpoint of AU Is the same point as the midpoint
,,] HI I)
X Prove:
THEOREM 77.23 If the vertices of a parallelogram are A = (0, 0),
B = (a, 0), C =; (a + b t c), D = {b, c), then the parallelogram is a
rectangle if and only if b = 0.
(You must prove ( 1) if ABCD is a rectangle, then h = and (2) if b = 0, then
ABCD is a rectangle.)
4. Prove:
THEOREM 11.24 The diagonals of a rectangle are congruent,
(Let ABCD be the given rectangle. V$e Theorem 11,23 and prove
AC m BD.)
502 Coordinates in a Plane Chapter 1 1
5, Justify the steps in the proof of the following theorem.
THEOREM 11,23 If the diagonals of a parallelogram arc congruent,
then the parallelogram is a rectangle.
Proof: Let A = (0, 0), B = (a, 0). C = {a + b, c), D = (b t c) be the
vertices of the given parallelogram. What theorem justifies our writing
C = (a + fc, c)?
We have
Therefore
and
Therefore
AC = BD.
(AC)* = (BD) 2 ,
(AC) 2 = (a + &)» + ^ 2 >
(«Z))8 = {&  o^ + c 2 .
( fl + b) 2 + c 8 = (J>  «)2 + c a .
Simplifying,
a 2 + 2<ii> + L* 4 c 2 = b 2  2afe f a 2 + c\
2ab= 2ab t
and
4«/> = 0,
Staee 4a 7^ 0* we can divide Ixith sides of the last equation by la, ob
taining h = 0. Therefore ABCD is a rectangle.
6. Justify the steps in the proof of the following theorem .
THEOREM 11.26 A rectangle is a square if and only if its diagonals
are perpendicular.
(See Figure 1136.) By Theorem 11.23
we may write A = {0, 0), B = (a, 0),
C = (a, c), D = (0, c) for the vertices of
the given rectangle. There are two things
to prove.
1. If AT X SB, then ABCD is a square.
2. If ABCD is a square, tlicm A~C _ WD,
Proof of 1: We arc given AT 1 UD. The slope of AC h r mid the
slope of #D is £—, Therefore
— a
— a
11.9 Proofs Using Coordinates 503
Therefore
c 2
 = _l ( fiS — fl a t an d f c f _ u , t
a*
But
c = BC and « = Afl.
Therefore BC = AB and A£GD is a square.
Proof of 2: We ate given ABCD is a square. Therefore
Afl = BC; \a\ = \c\ f and a = c 3 ,
I lit r»j be the slope of AC and m> be the slope of BD. Then
Wi = — , 'Us = — . and mirn^ = — Q,
But
— = l s .so mim? a — 1,
Therefore W I BB.
7. Justify Ihc steps in the proof of the following theorem.
THFOBKV 11.27 If the vertices of a parallelogram are A = {0, 0),
B = (a, 0), C = (a + &, c), and D = {&, c), then the parallelogram is
a rhombus if and only if a s = &+ c*.
There are two things to prove,
1. If a 1 = b s + c* then ABCD is a rhombus.
2. If ABCD is a rhombus, then flS = 6 2  A
Proof of 1: We are given that a 2 = A* + e* so
a V&* + C*i
By tlie Distance Formula,
ADm y/& + A
Therefore AD = of. Also, AB = a\, Therefore AB = AD and ABCD
is a rhombus.
Proof of 2; ABCD is a rhombus, so
AD = AB
and
Therefore
and the proof is complete
504 Coordinates in a Plans Chapter 11
8, Prove:
THEOREM 11.2H If the diagonals of a parallelogram arc perpen
dicular, then the parallelogram is a rhombus.
(Hint: Let A = (0, 0), » = {a, 0). C = (a + b % c), and D = (6, c) be
th e ve rtices of the given parallelogram. l,et mi and m% be the slopes
of AC and BD, respectively. Using Theorem 11.27 of Exercise 7 and
mi • mt = — I , show that a 2 = b 2 + cK)
9. Recall that a trapezoid is a quadrilateral with at least one pair of parallel
sides which are called the bases of the trapezoid. The other two sides
are called the legs of the trapezoid. The segment joining the midpoints
of the legs is called the median of the trapezoid. Prove the following
theorem.
THEOREM 11.29 The median of a trapezoid is parallel to each of the
bases and its length is onehalf the sum of the lengths of the two bases.
[Hint: Let A = (0, 0). B = {2a, 0), C = (2a\_2c), and_D a (2b f 2c) be
the vertices of the given trapezoid. Then AB and CD arc the parallel
bases. I*et E and F be the midpoints of AD and BC r respectively. Show
that EF I AB t ET  CD, and that
EF = fy\B + CD).)
10. A trapezoid is isosceles if its legs arc congruent (See Exercise 9.) Prove
the following theorem,
THEOREM 11.30 A trapezoid is isosceles if its diagonals are
congruent
1L Prove:
THEOREM 11.31 If a line bisects one side of a triangle and is parallel
to a second side, then the line bisects the third side of the triangle.
12. Prove:
THEOREM 11.32 The midpoint of the hypotenuse of a right triangle
is equidistant from the vertices of the triangle.
(Hint Let A = (0» 0), B m (2a, 0), and C = (0, 2fc), where a, b are
positive numbers, be the vertices of the given rigjht triangle.)
13. Complete the proof of the following theorem.
THEOREM 11.33 The lines which contain the altitudes of a triangle
are concurrent. (Their common point is called the ortbocculer of the
triangle.) (Sec Figure 1137),
Froof: Let A = (a, 0), B = (b, 0), C = (0, c)> and suppose that a < b t
< c as shown in Figure 1 137. I jCI A', B\ C be the feel of the per
11,9 Proofs Using Coordinates 505
i
kj
sm
t
/
\ \
* \.mo) i.
* *
t
t
/
Ugam IU1T
pcndfculars from A, B, C to BC, CA, AB, respectively. You arc to prove
that AA', BB', CC' arc concurrent at some point F{xi, y{), (Note that
(HO, 0) is the origin.)
The slope of BC is
b
c0
0h
•v — r ■* r
— — y. Since .'LA' 1 EC, the slope of
A A' is — (Why?), Using the point slope form of an equation, we may
express A A' as +_+ *
AA' = {(x,t 7 }:i/=:^a;.}.
* — > 4 — >
since (a, 0) is a point on AA'. We may express CC as
CC={(x 1 tj) :* = <)}.
Now write an equation for BB\ Let P(*i, 1/1) be the point of inter
section of lines AA', CO, and let F[x2, </?) b* f he point of intersection
of lines BB', CC, Solve the equations for AA', CC "simultaneously"
to find P[xi, t/i) an<l solve the equations for BB', CC simultaneously to
find P(x 2 , ?/2) Show that P = F and hence the lines AA\ BB'. CC are
concurrent at P.
14. Prove Theorem 11.18 by use of definitions, postulates,, and theorems
studied before this chapter: that is, without the use of coordinates.
(Perhaps you did this in an exercise of Chapter 10.)
15, Use Theorem 11,38 and theorems, definitions, and postulates studied
before this chapter to prove Theorem 11.29 without the use of
coordinates.
16* challenge Pitoni.KM. Prove Theorem 11.19 without die use of
coordinates.
506 Coordinates in a Plane Chapter 11
CHAPTER SUMMARY
In this chapter we used the idea of a coordinate system on a line to de
fine an .Tj/coordinate system in a plane. We showed that there is a oneto
one correspondence between the set of all points in a plane and the set of all
ordered pairs of real iwmlwrs. The key theorems in this chapter are:
THEOREM 11.4 If F = (x u yi) and P? = (** y 2 ) are any two
points in an .vj/plane, then
PiPs = V(*i  «a) B + (m  W<
THEOREM 11.5 If P = (x h ijt) and Q = (x 2 , ij 2 ) are any two distinct
points in an jt]/planc, then the midpoint X! of PQ is the point
\ 2 ' 2 r
TlIEOR EM 11.7 UA(xutfi) and Bixt, ys) are any two distinct points,
then
AB = {x, y) : x = x t + k(xt  *i), y = y, + %o  y^fcisreal}.
If jt is a real number and if F = (x, y) where ar = Xi + fe(*2 — Xi),
u = yj  %a  yi)» &<»»
AP = *(ArJ) and P^AJ? if it > 0;
AP = _*(AJ9) and P£oppAR if Jt < 0
The formula in Theorem 1 1.4 is called the DISTANCE FORMULA.
The formula in Theorem 11*5 is called the MIDPOINT FORMULA.
The equations x z= x% + k(x$ — Xi) and y = tji ■+ fe(j/ 2 — #i) in The
orem 11.7 are called FARAMETRIC EQUATIONS for the line A~S, and
k is called the PARAMETER. We proved that the converse of Theorem
11.7 also holds; that is, if tt, h, a, d are real numbers, if b and d are not both
zero, and if
S = {{x* y) '. x = a + bk, y = c + <&, fc is real),
then S is a line.
We defined the SLOPE of a non vertical line to be the slope of any one
of its segments. We defined the slope of a non vertical segment with end
points Pi(*i, t/i', 'Wt**! Vs) *° l** 5 ■ We showed that two non vertical
lines are parallel if and only if they have equal slopes and that two oblique
lines are perpendicuilar if and only if the product of their slopes is — 1. Slope
is not defined for a vertical line, but the slope of a hori7.ontal line is zero.
We proved that if P(*j, y{f and Psfra, y?) are any two distinct points
on a nonvertical line, then
*  *i _ V  tfi
*a  *i J/2  y\
Rsvlew Exercises 507
is an equation of the line. We called this form of equation the TWOPOINT
FORM for an equation of a line. We proved that a line with slope m and
passing through H x i> tf i) has an equation of the form
y  (/t = m(x  *i)
and called this form the POINT SLOPE FOHM for an equation of a line.
We proved that a line with slope m and {/intercept h has an equation of the
form
y s= mx + b
and called this form the SLOPE (/INTERCEPT FORM for an equation of
a line. We showed that an equation of a vortical line through ¥(%{, r/i) is
X — K\ and that an equation of a horizontal line through P{x\, iji) is y = y\.
Wc called the form Ax 4 By + C = the CpENERAL FORM for an equa
tion of a line.
Finally, we showed how coordinates could be used to construct proofs
for some geometric theorems and observed t hat, in some cases,, proofs using
cord (nates are easier than those using previously established definitions,
postulates, and theorems,
REVIEW EXERCISES
Graph each of the sets indicated in Exercises 110.
1. {{*, y):x = 5, < y < 5}
2. [(*f):r 3, 2<C*<7}
3. {(x, y) : x = 1 + 2Jfe, y = 2 + 3&, k is real}
4. ((* y) : x = 1 + 2*, y = 2 + 3*. k £ 0}
5. {(x, y) : x = 2 + fc, i/ = 1  2k k < 0)
6. {(a*, t/) : x = *, y = 3  2k, 2 < k < «}
7. {{*,«,) : l<x<4or2<y<5}
8. {(x, y) i 1 < s < '1 and 2 < y < 5}
9. {(x,y):y>2}
10. {{x, y) : y = x + 5}
In Exercises 1113, the endpoints A and R of a sequent AB are given. Find
(a) the slope of AM (in lowest terms), (b) the midpoint of SB, and (c) the dis
tance AB.
11. A = (2, 5} and ii = (2,3)
12. A  ;L 3) and B = (2, 9)
13. A = (2, 5) and B = (7, 7)
14. A = (2, 1) and B = (2, 3)
15. A = (3, l)andB= (7,1)
508 Coordinates in a Plane Chapter 11
16. Which segments in Exercises 1115 are
(a) parallel?
(b) perpendicular?
(c) congruent?
17. Write, in general form, an equation of the line which contains the points
P=(h ^3) and = (2, 9).
IS* Write, in general form, an equation of the line which has a slope of ^
and contains the point H = (—2, 1).
19. Prove that the line of Exercise 17 is perpendicular to the line of Exercise
18.
20. Write., in slope (/intercept form, an equation of the line with slope J
arid tyintercept 6,
■ In Exercises 2125, an equation of a line is given. In each exercise, (a) put
the equation in slope {/intercept form, (b) write the slope oi' the line, and
:::) write the x and {/intercepts of the line.
21. 3x + 2y m 12
22. 2x  y = 7
23. lox ~2lij = 7
24. x + y =
25. x — y =
86, Stow that A SKM is a right isosceles triangle if S = {3, 4), X = {  1, 3),
andM = (2, 1).
27. Given A = (1, 0), B = (4, 3), express AS using setbuilder notation and
parametric equation*.
28. Given A and B as in Kxerciso 27, express AB using setbuilder notation
and parametric equations.
29. Given A and B as in Exercise 27, find the trisection points of AB r
30. Write an equation of the vertical line through (2, 5).
31. Write an equation of the horizontal line through % 5).
32. Write an equation of the line through (3, —7) and parallel to the line
with equation y = 3x f 5,
33. Write an equation of the line through (3, —7) and perpendicular to the
line with equation tj = 3x + 5.
34. Given the line with equation 5x  % = 00. write the equation of this
line in slope {/intercept form.
35. Given the line of Exercise 34, write the equation of this line in intercept
form.
36. Given p = {(x, y) : x = 3} and q = {(x, y) : 2x = 6}, explain why
Revww Exercises
505
Exercises 3747 refer to the rectangle PQRS whose vertices are
P = (1, 4), Q = {5, 4), R = (5, .3), S = (1, 3;.
37, Fmd die midpoint oi PR.
38, Find the midpoint of SQ.
39, Show that PQ = $R r
40, Show that PR = SQ.
41, Write parametric equations for PQ,
42, Write parametric equations for PR.
43, Find A on Pit such that PA = 4PR.
44, Find B on PR such that PB = fPR.
45, Find C on opp M such that PC = J/»«.
46, Write parametric equations for the line through 5 and perpendicular
to PR.
47, Write parametric equations for the line through R and parallel to SQ.
48,
Let die trapezoid ABCD have vortices A = (0, 0), B = (2a, 0),
C = (2(1, 2c\ and D = (2/?, 2c), as shown in the figure. Let a, b, c, d be
positive numbers such that b < d < a. Let E and F be the midpoints of
AD and BC, respectively. Let EF intersect AC at P and SD at {). Show
that P is the midpoint of AEL that Q is the midpoint of ED, and that
P£ = {{AB  CD).
P(M),2e) C(2d,2e)
£(2b,0)
49. Prove, using coordinates, that the set of all points in a plane equidistant
from two given points in the plane is the perpendicular bisector of the
segment joining the given points. (Hint: Let the two given points in an
xyplane be A = (— a r 0) and B = (a, 0). Then the {/axis is the per
pendicular bisector of AB in the given .tryplane.) There arc two things
to prove.
(a) If P(x, y) is on the yaxis, then AP = PH.
(b) If AP = PR and P is in the .typlane, then F is on the yaxis.
50, challenge piiohlum. Prove that the urea S of a triangle whose ver
tices are A = (x\ t tj\), B = (X2, y2), and C =a [xz, y$) is given by the
formula
s = i^i^a + *»03 4 * 3 yi ~ *lft  x *y*  *fc§fll*
Bradley Smith/Photo Researcher!,
Coordinates
in Space
12.1 A COORDINATE SYSTEM IN SPACE
In Chapter 3 we introduced the fundamental idea of a coordinate
system on a line, or a line coordinate system, or a onedimensional co
ordinate system, as it is sometimes called. Tn Chapter H, we defined a
coordinate system in a plane and called it an *j/coordinate system. An
anycoordinate system is a onetoone correspondence between all the
points in a plane and all the ordered pairs of real numbers. Each point
has two coordinates. An ^coordinate system Is a twodimensional co
ordinate system. In this chapter we introduce the idea of a coordinate
system in space and called it an xi^coordinate system. In this system
each point of space is matched with an ordered triple of numbers. It is
•a threedimensional coordinate system.
Let a unit segment and die distance function based on it be given.
All distances will be relative to this unit segment unless Otherwise
indicated.
Let OX and OY be any two perpendicular lines and let OZ be the
unique line that is perpendicular to each at their point of Intersection,
Let J, J, K be points on OX, OY, OZ % respectively, such that
01 = OJ = OK = 1.
512 Coordinates in Space Chapter 12
On OX there is a unique line coordinate system with O as Origin
and / as unit point. On Or there is a unique line coordinate system
with O as origin and / as unit point. On O'Z there is a unique line co
ordinate system with O as origin and K as unit point. We call these
coordinate systems the xeoordinate system on OX, the {/coordinate
system on OY t and the ^coordinate system on Sz. We refer to OA\
OY, OZ as the xaxis, the (/axis, and the *axis, respectively We refer
to them collectively as the coordinate axes.
The plane containing the x and y~axe$ is called the xy plane, The
plane containing the x and saxes is called the j/zplane. The plane con
taining the y and saxes is called the r/zplane. We refer to these three
planes collectively as the coordinate planes.
From die theorems regarding parallelism and perpendicularity in
Chapter 8 it should be clear that all lines parallel to the .saxis arc per
pendicular to the K^plane, that all lines parallel to the (/axis are per
pendicular to the xzplane, and that all lines parallel to the Xaxis are
perpendicular to the 1/2plane, It should also be clear tit at all planes
parallel to the xyplane are perpendicular to the zaxis, that all planes
parallel to the xzplane are perpendicular to the [/axis, and that all
planes parallel to the i/splane are perpendicular to the xaxis.
Figure 121 suggests the x, i/, and ^rcoordinate systems, Hie
parts of tlie axes with negative coordinates are shown by dashed lines.
They are not "hidden" from view by the coordinate axes in the figure.
However, they are hidden from view by die coordinate planes, "llius
the negative part of the xaxis is behind the i/spkne; the negative part
of the c/axis is hidden by the aceplane; the negative part of the caxis
is hidden by tlie X(/planc, In drawing pictures you should use your own
judgment about whether a dashed segment is better than a solid one.
Figure IS 1
12.1 A Coordinate System in Space 513
In Figure 121, the positive parts of the x, y 3 and £axes are ar
ranged as the thumb, forefinger, and middle finger, respectively, of a
right hand when it is held as suggested in Figure 122* An .ti/5coordi
natc system with axes oriented in this manner is called a righthanded
coordinate system. If the unit points on the axes are selected so that
the positive parts of the x, y, zax&s con form to the orientation of the
thumb, forefinger, and middle finger of the left hand when the thumb
and forefinger are extended and the middle finger is folded, the xyz
coordinate system is called a lefthanded coordinate system. The fig
ures in this book arc for a righthanded system.
Figure 132
There is an xycoordinate system in the xyplane determined by
O, /, / as in Chapter 1 1 . This system is a onetoone correspondence
between the set of all points in the xyplane and the set of all ordered
pairs of real numbers. Similarly, there is an xscoordmate system in the
xsplane and a yscoordinate system in the y^plane.
T^et F be any point in space.
Figure 123 shows P as not
lying on any of the coordinate
pianos. However, the following
discussion which leads to the
definition of an xf/zeoordinate
system applies to any point.
For special positions of P some
of the labeled points thai are
distinct in Figiire 123 may
not be distinct. For example,
P and F IU may be the same
point.
Figure 123
514 Coordinates in Space
Chapter 12
Let P w P«, V vt be the projections of P on the xt/plane, the xsplane,
and the i/splane, respectively. Iet a XUt a gs , a^ be the planes through
P and parallel to the xyplane, the xsplaiic, and the i/zplane, respee
lively. Then a rv contains P t P rs , and P us ; fife* contains P, F m and F m %
and %, contains P, P Mt and /',
••..,.
Let @ n py, () t be the points in which a lw , a X!t a^ intersect the
xaxis, the yaxis, and the saxis, respectively. We are now ready to set
up an .rr/seoordinate system.
The ^coordinate of P is the ^.coordinate of Qy, the ycoordinate of
P is the ^coordinate of Q^ the scoordinate of P is the ^coordinate of
Q z . Wc write
P = (a, 6, c) or P(a, Z? } c)
to indicate that the *, y> scoordi nates of P are a, b, c, respectively.
If (a, h } c) is any ordered triple of real numbers, then there is one
and only one point P such that
P=P( a ,b>c),
It is the intersection of three planes, one parallel to tbe xsplane and
cutting the *axis at the point whose xcoordinate is a, etc. 'The cor
respondence between the set of all ordered triples of real numbers
and the set of all points is a onetoone correspondence. For if {«, b, c)
and (rf, e t f) are different triple*; of numbers, then one or more of the
following inequalities must hold:
a^d, 6#e, c=£f.
Suppose, for example, that a =£ tl Then P(«, b t c) and R(d> e, f) He in
distinct planes parallel to the i/zplane and therefore P^R.
Definition 12.1 Given an taxis, a (/axis, and a zfaxis, the
onetoone correspondence between all the points in space
and all die ordered triples of real numbers in which each
point P corresponds to the ordered triple (a, h, c) where a, b, c
are die Xh, y, ^coordinates, respectively* of P is die xyz
coordinate system,
Since the correspondence between points and triples is onetoone,
a system of names for the triples is a suitable system of names for the
points. Thus (5, 6 r 3) is an ordered triple of real numbers. It is also a
point. It is the point whose x, y f and scoordinates are 5, 6, and —3,
respectively.
12.1 A Coordinate System in Space 'j l :■
Example 1 If
A = {(*, y, z) : x = 2. tj  3, * = 4},
then
A m {(2, 3, 4)},
that is, A is the set whose only element is the point (2, 3, 4).
Example 2 If
B={{x t t,,z):x = 2.y = 3},
then J3 is the set of all points (a; y, 5) such that the xcoordinate is —2
and the ycoordinate is 3, that is, £ is the line through ( — 2, 3, 0) and
parallel to the .7 axis.
Example 3 If
C = {{*, y, z):y = 7},
then C is the set of all points whose ^coordinate is 7, that is, C is the
set of all points that are 7 units to the right of the acsplancj hence C is
the plane that is parallel to the xzplanc and 7 units to the right of it.
Example I
D = {(*, y, z) : x = 2 and y = 3}.
The use of "and" !n this setbuilder notation is the same as the use of a
comma between statements of conditions in a setbuilder notation. In
other words,
D = i(x i y i z}:x = 2,y = 3}
is the set of all points (2* 3, z\ that is, the line parallel to the saxis and
passing through (2, 3, 0).
Example o
E = {(x, y, z) : x = 2 or y s 3}.
Clearly* E is not the set D of Example 4. If you do not understand
dearly the distinction between the use of "and" and "or" in a set
builder notation, you shoxild review Chapter 1. E is the set of all points
with ^coordinate 2 or ^coordinate 3, hence the set of all points lying
either in the plane x m 2 or in the plane y = 3. Tn other words, £ is the
union of two planes, each parallel to the saxis, one of them parallel to
the yxplanc and 2 units in front of it, the other parallel to the xzplane
and 3 units to the right of it, assuming that the axes are as in Figure
121,
516 Coordinates in Spac* Chapter 12
Example 6
F = {(x, y,z):x=l + Zk t y = 2^ 3k t k is real}.
Now
{(*, y) : x = 1 + 2k, y = 2 + 3k, h is real}
is a line I in the xt/planc with slope  thai pisses through (1> 2). The
setbuilder notation for F places no restriction on the ^coordinate.
Therefore, if {x, y) is any point on I and z is any real number, then
(x, y, z) is a point of ft Conversely, if (*, y, z) is any point on F, then
(x, y) is a point of /. Think of /' as the set
{(a; j/,2);a;=l + 2*,j/ = 2 + 3k, z = 0, k is real}.
Then V is the same set as I If we think of a point of I, we think of it as a
point (x, y) in an xyplane. If wc think of the same point as a point of /'»
we think of it as a point (x t y, 0) in an xipspacc. Although the names
of the points are different, the sets I and I' are the same. The graph of
the set V is shown in Figure 124,
Figure 124
Now V is a part of ft Indeed, F is the set I' and all points directly
above it and all points directly below it. It may be helpful to think of it
as the union of all linos parallel to the *axis that pass through a point
of V. But any way you look al F t you really have not seen it unless you
recognize it as a plane, the piano through /' and parallel to the *axis.
Example 7
G = {(x t y,z):z<3}>
G is the set of all points (*, y, z) whose zcoordinate is less than 3. Now
z = 3 is an equation of the horizontal plane, say a, that is 3 units above
the xc/pkne. Then C is the halfspace that is the underside of a.
12.1 A Coordinate System In Spact 517
8
H=[(x t y 1 z):l< i z< 3}.
H is a slab or zone of space bounded by two horizontal planes. H is the
union of the horizontal planes with equations 5=1 and z = 3 and
all of the space that lies between them.
Example 9
I 8 {(*, tj,z):x* + y* = 25 t z = 0).
1 is a cirde in the xyplane. Its center is the point (0, 0, 0). Its radius is 5.
Example 10
/ is a solid right circular cylinder. Its axis is a part of the saxis. Its
radius is 5. Its lower base lies in the xi/plane. Its height is 6.
Example 11
K = {(*, y, z) : x = 2 and x = 3}.
There is no number x such that x = 2 and x = 3. Therefore there is
no point (x, \j, z) such that x = 2 and x = 3. Therefore K = 0, the
null set.
Example 12
L = {(*, y, z) : x*  3* + 2 = 0}.
Now x>  3* + 2 = is true if and only if (x  l)(x  2) = 0; hence
if and only if x = 1 or x = 2. Therefore L is the union of two planes,
each parallel to the t/zplane, one of them 1 unit in front of it, Ihc other
2 units in front of it.
EXERCISES 12.1
In Exercises 135, a set S of points is given in setbuilder notation In each
case, descrilw the set S in words, assuming that the axes appear as in
Figure 12 1. {Hint: For Exercises 20, 21. and 22, compare with Example 12.)
L $={{x t y.z):x = 0)
%S={(x,y.z):y = Q}
3. S = {(x, y. z) i z = 0}
4. S= ((x,y,2):x = 0,y = 0)
518 Coordinates in Space Chapter 12
5. S = {<x s y, 4 19 m %* = 0)
6. S= {(x,y, £ )ry = 0»z = O)
7. S = {(x, y, s) : x = Q, y = Q, s = 0}
8. S={{x,y, S >;x=l,y = 2,z=2}
9. S = {(*, y, z) : x = ly = 2, ] < * < 5}
10. S={(*,y,z) : a: = I, 1 <y^3,* = I)
It S = {(x, y, 2) : * £ 1, y = 3, z = 1}
12. S = {(*, y, s) : y < 3}
13. S = {{x, y, z) : x = 3, tj < 3}
14 t S = {(x, y, 2) : 1 <, x < 2, 3 < y <, 5, 1 < z < 3}
15. S = {{*, £/, z) ; x £ 0> y > 0, * > 0}
16. S = {{x, y, z):x> 370}
17. S = {(x, y,z):x = ly = 2)
18. S ss {(x, y, jj) ; x = 1 or J/ = 2}
19. S= {{x,y r z) : x = 5orx = 7}
20. S = ((x, y, 2) : s* + 3s + 2 = Oj
21. S = {(x, y, z) i & + 2* + 1 = 0}
22. S = {(x, y, s) : x 2 = 16}
23. S={(x,y,s}:x 2 + y 2 = 25)
24. S = {(x, y, z) : x 2 + y* = 5, *  5}
25. S= {(x,y, s ) :i/ = * + 3,5 = 4}
26. S = {(x, y, s) : tf s x}
27. S = ((x, y, *) : z = y}
2S. S = [<x, y, z) : 3x + Ay = 12, x £ 0, y > 0, 1 < 2 < 2}
29. S={(^, / ^):^ll == J^L? tS = 0}
30. S = {(x, y, z) ■ 2±1 = £zL. y = }
3L S = {(x r j/,*):.t = 1 + 2k, y = 2  ft, z = 0, 1 < Jfc < 2}
32. S = {(x, y, z) : x = 1 + 2*, y = 2  Jfc, 1 < * < 2)
33. S = {[x, y, g) : x= + y* ^ 2 5, s = 3)
34. S = (fx, y, z) : y 2 + z* < £5. x = 3)
35. S = ({x,y,s):x#0.y.^O\s ? feO}
■ In Exercises 3650, use set I milder notation to express the set 5.
36. 5 is the xyplanc,
37. S is the xsplane.
35. S is the j/*p1ane.
39. S is the xaxis.
40. S is the yaxis.
12.2 A Distance Formula 519
41. S
42. S
43. S
44. S
45. S
46. S
47. S
48. S
49. S
50. S
is the plane through (2, 6 f 7) and parallel to the yzplane.
is the plane through (2, 6, 7) and parallel to the xsplane.
is the plane through (2, 6, 7) and parallel to the xypfane.
is the line through (2, 6, 7) and perpendicular to the t/2plane.
is the line through (2, 6, 7) and perpendicular to the xsplane.
is dio line through (2, 6, 7) and perpendicular to the xyplane.
is the line in the xr/plane which contains (2. 3, 0) and (3, 7, 0),
is the ray AB, with A = (5, 3, 0) and B = (4, 6, 0).
is the segment Fp, with P s (2, 1, 0) and Q = (0, 7. 0).
is the segment RS, with R = (2, I, 1) and S = £0, —1,1},
12.2 A DISTANCE FORMULA
In this section we develop a formula for the distance between two
points expressed in terms of their coordinates. We begin by consider
ing an example.
Example 1 Let A  (3, 2 t 1) and B = (5, 4, 2). (See Figure 125.) I .el
A a = (3, 2, 0), B t = (5, 4, 1), B 2 = (5, 4, 0), B 3 = (0, 4, 1), and
B* = {0. 4, 2). Then AA2B2B1 is a rectangle and ABi = Ajfiz. Also
BBiBzB* is a rectangle and B t B = BsB 4 ,
Bjl Figure 1S5
To find A2B2 we use the Distance Formula and the x and ^coordinates
of A 2 and 13* In the xyeoordinalc system. As = (3, 2) and B 2 = (5, 4).
Therefore
A 2 B 2 = V(5  3)* + (4  2) a = \/8 = 2V2.
520 Coordinates in Space Chapter 12
To find B3B4 we use the Distance Formula and the y and subordinates
of B3 and B+ In the ^coordinate Systran B 3 = (4, 1) and B 4 = (4, 2).
Therefore
B S B^= V(4  4)z + (1  2)2 = VT= L
Now BBi is perpendicular to the xypkne. It is also perpendicular to
every plane parallel to the ayplane. In particular, S?i is perpendicular
to the plane with equation !■! Then mh & perpendicular to every
* — * ■ — *
line in that plane through tfj. Therefore BB t _L AB t and AABiB is a
right triangle. It follows from the Pythagorean Theorem that
{AB)* a (AB l )2 + (B^)*
(AS)* = (A 2 i3 2 ) 2 + (£ 3 Bi) 2
(AB)* = (V§) 2 + I 2
(AB)S = 8+1 = 9
AJ3 s 3
Following the procedure used in this example we shall prove the
next theorem.
THEOREM 12.1 {Distance Formula Theorem) The distance
between F{x lt y it z{) and Q{x& y% z 2 ) is given by
PQ = Vta  *i)» + (sf!  yir + (*  *)*.
jftoo/* Let P(*i, y lp si) and ()(x a , 1/2, afe) be given. Let
P x = (xi, yi f 0)
Qi = (*fe gfe 21)
Qz = (*a, y 2 , 0)
<?a = (0, y 2 , xi)
Figure 126 shows these seven points as distinct points. Depending on
die values of the coordinates, the points may turn out not to be distinct.
For example, if zj = 0, then P = P x .
We proceed to find PQ in terms of the coordinates of P and Q. If F
and Q\ are distinct points, then they lie on a plane parallel to the
u/plane and PQ ± is parallel to the xyplane. Therefore
(1) PiQtQiP Is a rectangle,
or (2) P = Ft and ft = Q 2t
or (3) F = Q L and P, = £ z .
12.2 A Distance Formula 521
Figure 120
In all three cases it is true that
(FQ 1 f = ftQtf = fa  *i)» + (y*  yi) 2 
Similarly, (I) QQ1Q3Q4 is a rectangle, or (2) Q = Q 4 and £i = Qz,
or (3) Q = pi and p 4 ■ Qn In all three cases it is true that
If 9 = p Xi then % = s 1? (sz — zi) 2 = 0, and
Hf = (PQtf = {x 2  xj* + (tfz  y,) 2 + (z 2  ft) 2 .
If P = p l5 then 12 = afe $2 = 5fo (*2  *i) 2 = 0. (1/2  gfo? = ft
and
(FQ)2 = (p!p) 2 = (ss  *i) 2 + (t/2  f/0 2 + (*b  *i) 2 
If Q ^ p x and P 7*= pi, then ^5, 1 FQi> mid it follows from the
Pythagorean Theorem tliat
(Pp) 2 = (FQtf + (WrP,
(Pp)2 = fe  X i)» + (y 2  yi) 2 + fo  s,} 2 .
We have now exhausted all cases. For if P and Q are any points and if
pi is related to P and Q as indicated in the first sentence of this proof,
thenP = Pi; or Q = Qn or P, Q, Q\ are three distinct points. We have
proved in all of these cases that
(Pp)2  fa _ Xl ) 2 + (tfz  yi) 2 + (^  *i) 2 
Therefore
pq = vte  xi) 2 + (ys  yi? + te  »i)*"
522 Coordinates in Space Chnptsr 12
EXERCISES 12,2
■ In Exercises 15, find the distance between Panel Q.
LPs (0, 0, 0), Q = (3, 4, 5}
2. F=(2, 1.7), Q= (2, 1.7)
3. P=(7, 8,1), p = (0 t fl,0)
4.P={0,0,5),@ = (3,4,0)
5. P=(a J,5),(? = (1, 1,6)
6* Use the Distance Formula to show that die following points are
collinear: A m (1,5, 2), B = (5, 7, 10), C = (2, 6, 6),
7. Use the Distance Formula to show tliat the following points are not
collinear: A = (0, 0, {)), B = {5, 3, 4}, C = (15, 10,  12).
a If A = (L 3, 2), H = (2, 7, 1), C as (2, 6,0), is A ABC a right triangle?
9. IfA = (5,3,2), B a (7,8, 1), C = (5,2, 3), is AABCa righttriau I '
10. Find the perimeter of &DEP if D = (0, % 0), E = (0. 6, 8).
F = (10, 6, 8).
■ In Exercises 1114, find tl*e distance between F{2. — 1, 7) and the given
plane a.
11. a = ((x, y. s) : x = 5} 13. a = {(x, y, =) : z = 0}
IS. a={{x,y,z):y = 0) 14. a : = {(x, y, z) : y = ~1)
■ In Exercises 1518, find the distance between the line / given by
I = {(x, y r z):x = 3, y = 4}
mid the given plane a which is parallel to it. (The distance between u line
and a plane parallel to it is not del ined in the book. 'Write a suitable definition
for tliis distance.)
15, a = {(r } y t z) : x = 0} 17. a = [(«, y, z) i ; ij a 6}
16. a — {(x, y s z) : i = —5) 18. a m {(x, y, s) : tj = 4}
■ In Exercises 1923, sketch a figure to represent the given set. (The graphs
will be threedimensional.}
19. A = {(*, y, s) : 1 < * < fi, £ < y < 4, < ;= < 10}
20. B = {(x, y, 5) ; 1 < a: < 2, 6 < y < 8. 4 < 2 < 6}
21. C = {(x,tj,z):l<x< 2, 10 < y < 12, < = < 10}
22. D = {{x, y> z) : 1 < x < 2, 14 <, y < 16, 4 < z < 5 or 6 < a < 7)
S3. £ = {(*, y, sj : 1 < x £ 2 t 18 < «/ < 20, < z < 10}
24, challenge problem. On one large sheet graph the union of the sets
in Exercises 1923.
12.3 Parametric Equations for a Una in Space 523
12.3 PARAMETRIC EQUATIONS FOR A LINE IN SPACE
Before studying Section 12,3 you should review the work on para
metric equations in Chapter 11 . By use of a proof similar to the one for
Theorem 1 1.7 the following theorem can be established, A proof of this
theorem is not included here.
THEOREM 12.2 If P(* 1t yu i) and Qix 2 , ua, z 2 ) are any two dis
tinct points, then
x = xi + k(x z  x\) t
W$ = (x, y, z) : y = tfi + k(tf2  yi)> and k is real
z = si ■+■ kizz — «i),
If R a {a, b, c) t where
a = xi + k{xz — x\),
h= yi + %j  tji),
c = zi + k(2t  *i),
is a point of FQ t then
It is the point P
R£PQ and PR = k * Pp
H € <W P£ and PR = Jfc ■ P@
Example 1 Given P = (0, 0, 0) and Q = (3, 1, 2), find the point R
on P$ such that PR = 3 ■ FQ and the point S on opp PQ such that
Solution;
P$ = {(x, U, z) : x = 3fc, y = *, * = 2fc, fc is real}
To get fl % set fc = 3. Then R = (9, 3, 6).
To get 5, set k = —3. Then S = (9, 3, 6).
Bampfe 2 Given A = (5, 1, 2) and £ s (7, 3, 2), find the mid
point M of AB.
Solution;
£jf = {(*, y, s) : x = 5 + 2fc, y = 1 + 2*, 2 = 2,kis real}.
The ^coordinate of A is 0; the fceoordiuate of B is 1 ; of M is £.
M = (5 + 2 ..^ 1 + 2  J. 2) = (6, 2, 2).
if
k = 0,
tf
Jt>0,
if
k<0.
524 Coordinates in Spact Chapter 1 2
Example 3 Given £ = (5. 7, 9} and F s (— 2, 3, —4), write para
metric equations for Et.
Solution: x  5  7Jt
y = 7 4k
s = 9  m
*£
Example 4 Given 11 = (5, 3, 14) and J = (2, 5,  1), write para
metric equations for lit
(1) with a parameter Jfc so that A: = at H and & = 1. at 1.
(2) with a parameter h so that /« = at I and fa = 1 at H.
(3) with a parameter t so that f = at 7/ and t = 10 at L
Solution:
(1) x= 5  7* (2) * = 2 + 7ft (3) x = 5  0.7*
y = 3 + Bk u = 5  8/i y = 3 + 0.8*
x= 1415* S=1 + I6ft z= 141.51
G < fc < 1 < « < 1 < f < 10
Example 5 Given J = (5,1, 3) and K s (7, —1, 5). find an equation
for the perpendicular bisecting plane of JK. A simplified equation is
desired.
Solution* Let a be the perpendicular bisecting plane of JR. Then
P(x> y, s) is a point of a if and only if
JP a EK,
tff)»  (IX)*,
(*  5)2 + (y  7)2 + (a  3)2 = (x  7) 2 + (y + If + (s  5)2,
x 2  lOx + 25 + y«  14?/ + 49 + s*  6s 4 9
= x 2  I4x + 49 + y 2 + 2y + 1 + s 2  10s + 25,
 lOx + 14x  Uy  %  6s + 10s
+ 25 + 49 + 9  49  1  25 = 0,
it  16y + 4z + 8 = 0,
x — 4ij r z + 2 = 0.
Then x — Ay + s + 2 = is "an equation of a" This means that
a = {(x, y, s) : x  4y + s + 2 = 0}.
12.3 Pwsroetrk Equations for a Line in Space 525
EXERCISES 12.3
In Exercises 14, find the coordinates of the point P satisfying the given
condition.
L A = (0 T 0, 0), B = (3, 3, 3). P€ 3$. AP = §>AB,
2. A = (0,0,0>,B = (3,3,3),P€ W a5,AP = * 3 *AB
3. A = (5, 2, 6). £ = (1, 3, 2), P C HA» AP = 2 ■ AD.
4. A = (3, 2, 5), fl = (5, 0.  1), P € XB t AP = FB.
In Exercises 510, A = (0, 0, 0), B s (6, 3, 12), C = (9, 18, 9).
5. Find D, the midpoint of J3fl
6. Find B, the midpoint of ACT,
7. Find F, the midpoint of A B.
8. Find Gi, the point that is twothirds of the way from A to D,
9. Find G2, the point that is twothirds of the way from B to E.
10. Find G& the point that is twothirds of the way from C to F.
11. Given S = {(x, y , z) : 3x + 4i/ + 5s = 50), which of the following
points are elements of S: A(5 t 5, 3), fl(G, 0, 5), C{ 1, 2, 12}?
12. challenge problem. Find tlie point on the Tiutis that is equidistant
from A{2, 3, 1) and B{7, 2, 1).
13. challenge problem. If S is the set of all points each of which is at a
distance of 7 omits from the origin. End an equation (in simplified form)
for S. What is a set of points like S usually called?
14. challenge pboblem. If T is the set of all points, each of which is at a
distance of 3 units from the point (3, —1.2), find an equation (in simpli
fied form) for T.
1.5. CHALLENGE PROBLEM. (Jiveil
S = {(Y, y, a) : & + f 4 z 2 = 125},
T = {(x, y, z):x = 5}, mid Q = (5, 0, 0),
prove that all points P in the intersection of S and T are at the same dis
tance from Q.
16. challenge problem. Let planes u and and points, O t A, A Ir B, B^
be given as follows:
/? = {(*. y, z) : j 4 y + s = 9},
O = {0, 0, 0), A = (2. 4.2). Ai = (6.0, 0),
B = (3, 3, 3). Bi = (9. 0, 0).
(a) Show that A and A i are points; of a.
(b) Show that B and Bi arc points of ft.
526 Coordinates « n Space Chapter 12
:V) Show that UA 1 ZKl.
(d) Show that UB 1 SSI,
(e) Find a point Aa in a such that A, A], A* are noneollinear.
(i) Find a point H* in /? s uch llial B» B lt Ba are non colli near.
(g) Show that UA ± AAg,
(h) Show that US 1 SSi.
(!) Show that DA 1 a.
(j) Show that DZJ 1 j8.
(k) SIiow that 0, .4, iB are colHnear.
(1) Find the distance between a and /i.
17. CHALLiiscjK iitonr.nM. I.et planes a and /3 and line / lie given as
follows:
a = {(x t y, z) : x + y + 2,? = 8},
= {(*, y, s) : x + y + 2* = 2}.
= [{x, i/, z) : x = y = ^H
(a) Find the point A in which / intersects a.
(b) Find the point B in which / intersects /?.
(c) Prove that A~B 1 a.
(d) Prove that AB _ fi,
(e) Find tlie distance between a and /?.
12.4 EQUATIONS OF PLANES
In Chapter II we used coordinates to write equations of lines. For
example,
3* + Atj +■ 7 =
is an equation of a line in an typlane* We say that
3x + 4y + 7 =
is a linear equation in y and t/. The equation
ax 4 ^ 4 G =
is called the general linear equation in x and y. It is natural to extend
this terminology and to call
ax + hy h cz 4 & —
the general linear equation in x, !/, z. In this section we show that, if an
xyz coordinate system has been set up, every plane has an equation of
the form
ax + by + '* + d —
12.4 Equations of Plane* 527
with «, h, & d real numbers and with a t b, c not all zero, and conversely,
that every equation of this form is an equation of a plane. What is the
situation ifa,h t and c are all zero? More specifically, what is the graph
of Qat + % + G*+I = 0? Of Ox + Qy + CMr + = 0?
We begin with several examples, assuming that an xy ^coordinate
system has l*ecn set up in space.
Example t IaH = (0, r 0), P = (3, 4. 6), and let a be the plane
that is perpendicular to UP at P. We shall exjyrefa a in terms of the co
ordinates of the points on it using setbuilder notation, Now
OP = (fr tf % z):x = 3k>y = 4*. z = 6*, k > 0},
I ei: O I h 1 1 1 1 point on OP that is twice as far from O as P is from O, To
find the coordinates of Q, set k = 2, then (J = (6, 8,  12), P is the
midpoint of UQ t and a is the perpendicular bisecting plane of U%).
The perpendicular bisecting plane of a segment is the set of all
points equidistant from the endpoints of the segment. Using the Dis
tance Formula, we find that the distance between O(0, 0, 0) and (x, y, z)
is
V* 2 + yf + a?
and that the distance between P{6, S, — 12) and (x t y t z) is
V(*  6)2 f {y  8)2 + ( S + 12)3.
Therefore
« = {(X, y, Z) : # + 7 + ^
= v'«*  8)* + {y  8) 2 + (* + 12) 2 }.
a = {(r, p, c) : s* + y 2 + s 2
= x 2  12% + 36 + t/2  101/ + 64 + z 2 + Uz + 144) f
« = {(*, y. s) : Hx + 1%  24s  244 = 0},
« = {(*. y» s) : 3* + 4y  6*  61 = 0).
Example 2 Let O = (0, 0, 0% P = (3, 4,  6), and let /? be the plane
that is perpendicular to OF at O. Note that /? is parallel to the plane a
of Example 1 and that a. docs not contain O, but that fi does. Now
OP = ((*, r/ s z):x = 3k, y = 4Jt, z = 6k, k Is real}.
Let R be the point on opp OP such that OH = OP To find the coordi
nates of R, set k = — 1, then K sb (—3, — 4, 6), and O is the midpoint
of BP and $ Is the perpendicular bisecting plane of KP.
528 Coordinates in Spac* Chapter 12
/? is the set of all point 4 ! (x, y, z) each of which is at the same distance
from (3, 4, 6) as it is from (3, 4, 6), Therefore
P = (fe .'/> A = V(«  3? + (y  4)» + fr + 6)2
= ^ + 3)> + (y + 4)* + («6)S} t
fi = {(*, y, «) ■ : x s  &x + 9 + i/2  8y + 16 4 z 2 4 12s + 36
= x 2 + 6* 4 9 + !/ 2 + 8y + 16 + s£  12s 4 36},
p a ((* //, 4 : 12*  16y + 24* = 0},
= {(*, t h Z) ; 3* 4 4y  62 = 0}.
ficample 3 Let y he the following set of points in .fysspace:
y = {(*, y, *) : *  2y + 3a4 = 0}.
We shall show that 7 is a plane,
Xote in Example 1 that an equation of a is
3* + 4y  Qs  6] = T
that the coefficients of x, y, z in this equation are 3, 4, — 6, that the foot
of the perpendicular from O to a is {3, 4, —6), that
32 + 4 s + (_e)*  61,
and that 61 is the negative of the constant term in the equation of a.
Tliis suggests that we transform the given equation of y,
x  2y 4 32  4 = 0,
into an equivalent equation in which the sum of the squares of the co
efficients is the negative of die constant term, that is, an equation
ax + hy + cz + d =
where a 2 4 b' £ 4 e 2 = —d,
Wc multiply through by k and then determine a positive value of
k so that the resulting equation of y has the desired property. We get
fcr  2ky 4 3Jb  4fc =
and we want
fe3 + ( _2fc)* 4 (3fc) 2 = 4* with fe > 0.
Therefore
# + 4/t 2 + 9ft* = 4k, 14*2 = 4^ fc m 2
12.4 Equations of Planas 529
Therefore
Note that
and that $ is the negative of the constant term of our **adj listed" equa
tion of v.
Taking a clue from Example 1, we suspect that y is the plane that
is perpendicular at (7,7,7) to the line through (0,0 4 0) and
{f t — * I }, and hence that it is the perpendicular bisecting plane of the
segment with endpoints (0, 0, 0) and (4, — J, 4^ ). To show that it is, we
let y' be that perpendicular bisecting plane and we show thai y = y>
Now y ' is the set of all points (st, y, z) each of which is the same dis
tance from (0, 0, 0) as it is from (^, — ■$» 4^ ). Therefore
Y m {{x, y, z) : V* 2 + U 2 + ^
= V(* *) 1 +(* + *)■ + <» V M.
/ = ft*. £f, «) : x 2 + if + ~ 2
Therefore y' = y t and y is a plane. Since
y = [{*, r/, s) : x  2y + 3s  4 = 0},
this proves that
x  2y + 3z  4 s
is an equation of a plane.
Example 4 Let 5 be the following set:
5 = {fr y,z):x2y + 3z = Q).
Since (0, 0, 0) is an element of 5, it follows that S is a set of points that
contains the Origin. We shall show that 5 is a plane.
Taking a clue from Example 2, we suspect that 5 is the plane that is
perpendicular at the origin to the line through the origin and (1, — 2, 3).
We suspect further that 8 is the perpendicular bisecting plane of the
segment with endpoints (I, —2, 3) and (— 1, 2, —3). Let 8' be that
perpendicular bisecting plane. Wc shall show thai 8' = 5,
530 Coordinates In Spae* Chapter 12
We have
5' = {(x, y, z) : Vfr  l) 2 + {y 4 2)* 4 (z  3) 2
= V(* + I) 1 + (y  2J 2 + ( Z + 3)*},
6" = (fr y t z) : x 2  2* + 1 + tp + Ay + 4 + z £  6z + 9
= x 2 + 2r + I + y 2  4y + 4 + s 2 + 6s + 9},
^ = {(a:, y, z) ; 4* f 8y  12z = 0},
5' = {{x > y > s):x2 !/ + 3* = Q}.
Therefore $* = 8, and 8 tea. plane.
We arc now ready for several theorems on linear equations in £, y t
and z,
THEOREM 12,3 Given an aryzeoordinate system, every plane
has a linear equation.
Proof: Let a plane « !>e given. We consider two cases.
Case I. O(0, T 0) does not lie on a.
Case 2. 0(0, 0, 0) lies on <*.
Proof of Case 1: Let P{a, b t c) be the foot of the perpendicular from
O to a. Then a, h, c are not all zero. Let Q = {2a, 2b : 2c). Then P is the
midpoint of UQ, and a: is the perpendicular bisecting pkne of UQ,
Then
a = {(*% y, z) : \/x* + y 2 + z 2
= V(*  2») 2 4 (y  2£>} 2 + (z  2c:) 2 },
0£ = {(%, y, z) : ax + by + cz = a 2 + b 2 + c 2 }.
Therefore in Case I, a has a linear equation.
Proof of Case 2: Let P{a, b, c) be a point distinct from O on the tine
perpendicular to a at O. Then a t h, c are not all zero. Let
Q = {—a. —b, — c). Then O is the midpoint of PQ and q is the per
pendicular bisecting plane of FQ. Then
«= {fcy,*) : Vfr  a) 2 + (y  *>) 2 + {z  c) 2
= >/(* + «) 2 + fr + *>) 2 + (* + «)"),
a = {(%, y, *) : x 2  2a* + a 2 + y 2  2fci/ + fe 2 4 z 2  2zc + e 2
= jc 2 + lax + a 2 4 y 2 + 2by + fe 2 + s 2 + 2sc + c 2 },
a m {{x, y, z) : ax 4 by + cz = 0}.
Therefore iu Case 2, a has a linear equation.
12.4 Equations of Plum 531
THEOREM 12.4 Given an ^coordinate system, the graph of
every linear equation
ax 4 by 4 ez + d = 0,
in which a f b t c,d are real numbers and a., b, c are not all zero, is a
plane.
Proof: Let an sy^coordinate system and an equation
ax J % + cz + rf = 0,
with a, A», c not all zero, be given. We consider two cases.
Case J. <*#().
Case 2, d = 0.
Proof of Case 1 ; Suppose d =£ and let
« = {(x, y, z) : ax + % f cs + rf = 0}.
Let /: l)e a number, not zero, to be specified later. Then
a = {(x, y T z):a f x + b'y f c'z + d' = 0} :
where a 1 = fa b* = kh, u' = kc, and ff = hd. Taking a clue from
Example 3, wc shall determine k t with Jt ^ 0, so that
d' = (a'}2  gtp  {&)*.
Substituting we get
kd= {kaf {kh)*  (kc) 2 .
Solving for k we gel
kd = Wo 2  Ffe2  k 2 c\
d = ka 2 kb* kc 2 ,
d = k(a2  fea _ ^
and this is our specified value of h Note that since a, b t c are not all
zero, then at least one of the nonnegativc numbers, a 2 , h 2 , c 2 , must be
positive, and a 2 4 h 2 + c 2 is positive. Hence the expression for k in
terms of a, b, c, cJ is mathematically acceptable since the indicated di
vision is division by a number that is not zero*
Checking wc find that
tt = kd = #(a2  68  c2) = — (Jfcu) 2  (Mi) 2  (fee)*,
d = (*>)*  (by  { C y.
532 Coordinates in Space Chapter 1Z
Let P = (a*, b\ c 1 ) an d Q = (2a' f Zb\ Icf). Let cr' be the perpen
dicular bisecting plane of OQ. Then
or' = {(x, y, 2) : V* 2 + y 2 + £
= >/<*  2a*)* + ( y _ 2bT + >  2c 1 ) 1 }.
= {*, i/, s) : *a + y* + & = * 2  4rf* 4 4(a')2
4 y 2  4b'y 4 4{fc') 2 + #  4c'.s + 4(c') a },
= (fo y, z) ■ u'x + //./ + <te  («7  (PT  WT  0),
= {(*, y, s) : a'* + fc'y 4 c's 4 tf  0),
a {{x, y, z) : kax + kby 4 fcos + kd — $}>
= {{ar, y, z) : ax 4 by + cz + d = 0},
and a' = a. Since & t h a plane, it follows that a is a plane, as we wished
to prove.
Proof of Cose 2; Suppose that r / = and let
a = {{x t y, s) : ox 4 by 4 cs = 0},
Then
a = {(x, y, z) : ox 4 by 4 cz 4 ii = 0}.
Let P s (a. 6, c)„ p = (— tt, — ft, — c), and let a' be the perpendicular
bisecting plane of TQ. We shall show that a' = a, and this will prove
that
ax + by + cz + d =
is an equation of a plane.
Now
«' = {fa y, z) \ Vfr  tt?» + (y  fr) a + fr  cp
= \/(x 4 u) 2 4 by 4 b)'* 4 {* 4 c) 2 },
a* = {(*, y } *) : a 2  2a* 4 0* +• y 2  2% 4 P 4 z 2  ^SJ 4 c 2
s x' 1 4 2a* 4 a 2 + y t + 2/?y
4 b 2 4 z 2 4 2zc 4 c 2 },
a' = {(x, y, z) : —4ax — 4by — 4cz = 0},
ex s {{r, */ t 2) ; ax 4 % 4 ra = 0} ,
"Therefore a' — a.
THEOREM 12,5 If a, b, c, J are real numbers with a, b, c not all
zero, then the plane
a = (£x, y, z) : ax 4 by 4 cs 4 d = 0}
is perpendicular to the line through 0(0, 0, 0) and P{a, b, c).
12.4 Equations of Plan** 533
Proof: If tt = 0, then as in the proof of Case 2 of Theorem 12.4 it
follows that o is perpendicular to the line Py y where
Q = (—a, —h, —c) and P = (a, b t c).
But O is the midpoint of PQ. Therefore Op = P<$ and a 1 OP.
If d t^ 0, then as in the proof of Case 1 of Theorem 12.4 S it follows
that a is perpendicular to the line OQ, where
Q = (2a\2b\2t/)>
Saw
0<$ = {{*, tj, z) : x = 2aft, y = 2b% z = toft, t is real}.
(We have used t as the parameter since k was used for another purpose
in the proof of Theorem 12.4*) Setting * = i, we obtain a particular
2k
point on OQ, namely ( ,  , ^~) = P(a, b t c). Since a is perpendicular
K ft 1\.
to Op, and since 5iy = OP, it follows that a is perpendicular to 5?.
Example 5 Find an equation of the plane perpendicular to OP at F if
O = (0, 0, 0) and P = (2 4 7, 0).
Solution; For every real number d, 2x — ly + d = is an equation
of a plane perpendicular to OP. (Which theorem or theorems of this
section are we using here?) To get a satisfactory equation we fix d so
tlial the plane contains P. Substituting the coordinates of P, we get
227(7) +d = and d = 53,
Therefore
2x  7y  53 =
is an equation of the plane perpendicular to Or at P.
Example ft Find an equation of the plane perpendicular to DP at O
if = (0, 0, 0) and P = (2, 7, 0),
Solution: For every real number d, 2x — 7tj + d = is an equation
of a plane perpendicular to OP. The origin O(0, 0, 0} is a point of this
plane if2*0 — 7*0 + 4 = 0, that is, if d = 0. Therefore
2x — 7y =
is an equation of the plane perpendicular to UP at O.
534 Coordinate* in Spact Chapter 12
Example 7 Find an equation of the plane perpendicular to OP at P if
O = (0, 0, 0) and P = (7, 4, 5).
Solution:
7* + 4y  5z + d =
7.7 + 4.4 _ 5(_5) + da
90 + d =
d a 90
7* + Ay  5s  90 =
Example 8 Given
o = {(*, t/, 2) ; 5sc  2y + 4a  8 = 0},
find u point P .such that « is perpendicular to the line through O(0, 0, 0)
and P.
Solution: P  (5, 2, 4).
Example .9 Given
a = {(x t y, *) ; 2k  7* = 0},
find a point P such that a is perpendicular to the line through O(0, 0, 0)
and P.
SbFtfftbn; F = (2 f 0, —7),
Example W Find an equation of the plane a that contains the points
O(0, 0, 0), A(l, 1, 1), and B(l t 7, 3),
Sofuft'cm; We want real numbers a, b, c y d with a, b, c not all zero
such that
a = {(x, y, z) \ ax + by + cz + d = 0).
Since O, A, B arc points of a, it must he true that
(1) a*O + ft*O + 0*O + <la>0,
(2) a •! + b'l + cl + d = (),
(3) a • I + b • 7 + &•{— 3) + 4 = a
From (1) we deduce that d = 0. From (2) and (3) we deduce by sub
traction that
1 6/;f4c = 0.
12.4 Equations of Plan** 535
Now neither b nor c can be zero. Why? To complete the solution we
take any number except for b and solve for the corresponding a and c.
Thus, if b = — 2, then c = — 3, a = 5, and
a a {(x, y, z) : 5x  2y  3* = 0}.
To check, substitute coordinates as follows:
O: 5*02030 = 000 = 0.
A: 512131 = 523 = 0.
B: 5*1273 (3) = 5  14 + 9 = 0.
Kxamplc II Find an equation of the plane that contains the three
points P(l, 5, 7), Q(l, 2, 4), R% 1, 5).
SoJti (ton;
(1) ax + % + ez + d =
(2) a • 1 + /?  5 + c • 7 + a" = 0, a + 5b f 1c + d =
(3) fl(l) + ft2 + c{4) + d = 0, a + 2fo4e + d =
(4) a 2 + fc * 1 + c  (5) + 4 = 0.2<t + &5c+<Z=0
From (2) and (3) we get (5), and from (3) and (4) we get (6).
(">;> 2a + 3& + lie =
(6) Za f b + c =
From (6) we get (7), and from (7) and (5) wc get (8).
(7) 9a + 3b + 3c =
(8) llo + 8c =
Take a = 8; then c = — 11. From (6) we get fc = 35 and from (2) we
get d = — 106, A satisfactory answer is
8* + 35y  llz  106 = 0,
Check:
P: 81435511*7
= 8 + 175  77  106 = 0.
Q: 8(l) + 352  11 (4)  106
= 8 + 70 + 44  106 = 0.
R: S2 + 351  ll(5) 108
= 16 + 35 + 55  106 = 0.
536 Coordinates In Span Chapter 12
THEOREM 12.& Consider the plane
a ={(*» if* x ) : ax + % + <* + d  °)'
If « = 0, a is parallel to the xaxis. If h = 0, a is parallel to the
yaxis. If c = 0, or is paralU] to the axis.
ftoo/v Suppose first that a = and d = 0. Then
a = {(*, y t z):by + cz = 0}
and for every real number x, a contains the point (jr., 0, 0). Therefore of
contains die araxis and is parallel to it.
Suppose next that a = and fi 7^ 0. Then
and since
a = {{x, y, z) : by + cz 4 d = 0}
bO + cO + d^O,
it follows that no point (x, 0, 0) is a point of a and henec that the xaxis
is parallel to id
Similarly, it may be shown that if b = 0, men a is parallel to the
ffaxis, and that if C = 0, then a is parallel to the saxis.
Examph 12 Let a, /?, y, 5 be given as follows:
a = {(*, y t z) : 3* + 4y  12 = 0},
/?={(*, i,,z):3x2z = 0},
7 = {(x, tj, z) : 5tj = 8},
* = {(*, i/, «) : 2*  5 = 0}.
(See Figure 127,) & is parallel to the saxis, a contains the line I in
the xyplanc widi x and (/intercepts 4 and 3, respectively, and every
line parallel to the saxis that passes through a point of I
Figure li7
12.4 Equations of Plane* 937
(See Figure 128.) /? is parallel to the yaxis; in fact, it contains the
i/axis. The plane /? contains the line I in the .replane that passes
through the points ((I } 0) and (2, 0, 3). It contains every line parallel
to the yaxis that passes through a point o( L
Figure 128
(See Figure 129.) y is a plane parallel to the araxis and the 3axis;
hence it is parallel to the .tsplane. y contains every point whose (/coor
dinate is 1.6 and every point of y has ycoordinale 1.6.
♦a
yml(x,y,i};5y>B\
Fignn 12W
(iSee Figure 1210.) 5 is a plane parallel to the yaxis and to the
>axis; hence it is parallel to the y£plane. The plane 5 is the set of all
points whose ^coordinate is —2.5,
* if*, y.*): 2* + *0f
Figure 1210
538 Coordinates in Space Cnap:e'12
EXERCISES 12,4
■ Let a plane « = {(*, y, s) : 4x — 3y + 180 = 0} be given. In Exercises
110, determine if the given point is a point of a,
1. (0, 90, 0) 6. (0, W, 137)
2. (45.0,0) 7, (45, 0,237)
3. (0,0,180) 8, (30.20,10)
4. (0, 0, 0) 9, (7, 69, j)
5. (0, 0, 37) 10, (100, 73± 6Xif)
B In Exercises 1 1 20, an equation of 8 plane is given. Find three numbers
Or b, c f not all zero, such that the given plane is perpendicular to the line
through the origin and P{a, b, c).
11. 3.r  7y + 14 = 16, fr ? nl=0
12. 2x  y + z  73 = 17. x  y + 10s =
13. 2y = 3x 18. x  y + 10s + 10 =
14. tj = 3x + 5 19. y + lOx  4s =
15. * = 3y — 3 20. x  y = s  1
■ In Exercises 2125, siin plify the given equation to an equivalent equation in
the form of the general linear equation ax + by + ex +■ d = 0. If one of
the coefficients is zero, Uiat term may be omitted in the simplified form.
Thus we accept 3* + 4y = in place of 3* + 4tj + Oz + = 0.
21. yx" + tf + z* = V(x l )»f (y2)' + (g 3) s
22. y '{*  2}* + (y  3?' + s' = \ /^T 2f+(y + 3)' + ^
23. V fx I) 2 + (r/  if ^UP e V (x 12j a + ( y  5) a f s*
M, y ^T4) g + (y + 4)» + (a T ffi = V* 2 ^ y» + ^
25. Vfx^T? + ~y*" + [5 +TP = y/{x  1)* + (y  1)2 + (z + 1)2
■ In Exercises 2630, find a linear equation in simplified form for the plane
that con tarns die diree given points.
26. (0, 0, 0), (8, a 0), (3, 0, 1) 29. (8, 0, 4), (0, 4, 3), (4, 4, 1)
21 (1, 3. 1), (0, 1, 1), (2, 1, 1) 30. (3, 0, 1), (9, 0, 3),
23, (5, 2, 3), (3. 6. 7), (0, 4, 6) (51, 10, 17)
B In Exercises 3140, ail equation of a plane is given. Find the coordinates
of its intercepts on the coordinate axes. If it does not intersect an axis, write
"none." The answer to Exercise 31 is given as a sample.
31. 2x  3s + 30 = xratercept; (15, 0, 0)
{/intercept: None
s intercept: (0. 0, 10)
12,5 Symmetric Equations tor a Un« 539
32. 3x + by  2x  20 = 37. tj = 3
33. *  y + s = 38. z = 4
34. 5*  4y + 100 = 39.x + y J e = 100
35. 2x  ij + s = 40. a : 4* y + = + 100 =
36. x = 6
In Exercises 4150, state whether ihc given set S is a plane, the union of two
distinct planes, n line, the union of two distinct lines, a set consisting of a
single point, or the null set.
41. S = {(x, y t z) : x = 2, y = 3, z a 4}
42. S = {(x, y. a) : x + y + s = 5J
43. S={{x t ,j,z):x+,j = 5}
44.S={(x t y):x+y = 5)
45. S = {(», y) : x = 5}
46. S = ((x, y, z):x = 5orx= 10}
47. S = {(x t y, x) : x + p + « = 10 and x + y + z = 20}
45. S = {(x, y, x) : x + y h z= 10 or x 4 y f s = 20}
49. S = {(x, y, z) : x + y + 2* = 10 and x  2y + 2z = 10}
30. S = {(x, i/, c) : x + y + 2s = 10, x  y + 2* = 10, 2 = 0}
12.5 SYMMETRIC EQUATIONS FOR A LINE
Since a linear equation in x and y represents a line in an xryplane
and a linear equation in x, y> z represents a plans in xj/zspacc, it is nat
ural lo ask how we might represent a line in terms of coordinates in
xysspaee and to do it without using a parameter. A clue to the answer
is provided by some of the exercises of Exercises 12.4. As usual, we
assume an xyxeoordinate system is given.
Example 1 If P = (2, l s 7) and Q = (5, 1, 3), then
The notation
x 2 y 1 x 7
52~ll : 37
is a short way of saying that
ill, y 1 and y 1 = * 7 .
5_2 11 11 37
540 Coordinates in Space Chapter 12
To see that the representation in terms of these equations is correct,
note that
*  2  y  J rt „i j/ 1 _ ~ 7
unci
52 11 1  1 .37
arc equations of distinct planes, one of them parallel to the saxis and
the other parallel to the itaxis. Note that both F and Q lie on each of
these planes. (Check this by substituting coordinates.) Hence FQ is
precisely the intersection of these two planes and the representation
we gave in terms of "symmetric" equations is a satisfactoiy one. Actu
ally, there is more symmetry in this situation than we have indicated
so far. It is natural to think of ry as die intersection of three planes
related to the three expressions equated to each other in the setbuilder
notation, Thus every point of FQ lies on each of the following three
planes, a t /?, and f, which are parallel, respectively, to the a>, y, and
2>axes:
* = [<* * *> ■■ fff = ^nrr)
Generali zing the situation in Example 1, we obtain the following
theorem.
THEOREM 12. 7 If P(x lt y it *i) and Qfa y 2i z 2 ) are two distinct
points with x% ^= 4*2, tj\ ^= tj>£, £\ ^ jgtg, then
FQ = { lx t y, z) : = * *— = 
I X2  Xl yz  Jfr 8£  «i
ftw/: Let
I »2 — 3^1 ^2 — 2»i )
I x 2 — xi y 2 — yi J
12.5 Symmetric Equations for a Lin* 541
Now P(xu j/i, 2i) is a point of a since — ^ = — — =
y* — yi z z ~~ *h
and Q(x 2 , ijt, %) is a point of a since  /a " yi = * 2 ~ Zl = 1.
^a  1/1 «■— »i
Similarly, we can show that P and p each lie on both /3 and y ,
Also, a, /i, y are three distinct planes. For a contains the point
(xi — X t tf], Zi} t whereas ft and 7 do not; fi contains (xi, 1/1 — 1, Zi),
whereas on and y do not: y contains (xi, yi,Z\ — 1), whereas a and fi do
not The — 1 here may appear to border on the magic. Actually it does
not For example, if Xo is any nonzero number whatsoever, then a con
tains (xi — Xo t yi, zi), whereas fi and y do not
Therefore (x, y, z) is a point of PQ if and only if it is a point of all
three of the planes a, fi, y, and hence the representation using sym
metric equations is a valid one.
THEOREM 12.8 If F(x u 1/1, *t) and Q(x 2 , y 2 , $§} arc distinct
points with xj ^= x 2 and y^ sfi y 2 > then
I X 2  Xl $2 IJl >
Proof;
X — Xi u — 5/1 ,
and z = «i
ia  xi ya  ^3
are equations of two distinct planes each containing both P and Q,
Therefore PQ is the intersection of two planes, and the representation
given in the statement of the theorem is a valid one.
Similar proofs may be given for Theorems 12*9 and 12,10.
THEOREM 12.9 If F(x u yu z t ) and p(x 2 , yu z^} are distinct
points with xi ^= x 2 and z± =£ ffg, then
i v*> y> «
X — Xi * — Zl
'}
Zi \
X 2 — X Z% — Zi i
1 HEO /t EM 12.10 Tf F(Xi , «/ 1, ?i ) and ()(xi , i/ 2 , z 2 ) are two distinc I
points with tji ffe 1/2 and zi 7^2. then
I yi  yi *2  zi )
542 Coordinates. In Space Chapter 12
Proofs of Theorems 12,11, 12.12, 12.13 arts assigned as exercises.
THEOREM 12 .1 1 If P(x lt y h ft) and Q(xi, y lt * 3 ) are two distinct
points, then
W = ((*. y.z}:x = xi and y = y t }.
THEOREM 12. 12 If P(x lt y x > zi) and Q{x x , y 2t Si) are two distinct
points, then
¥@ = {{*, y t z):x = xi and *= %).
THEOREM 12.13 If flfa, ^i> *i) and 9(* 2 , if! . a,) are two distinct
points, then
PQ = {(*, y, s) : y = yi and s = *i).
EXERCISES 12,5
■ In Exercises 15, a tine is given in terms of symmetric equations. In each
case, find the missing coordinates of P t Q t R so thai they will be points of the
given line.
m. ib m x e& m. m ). «{3. in. h )
"£3. [Em). «7. a, >, j!(n, s h )
1 5 3 2 J
nm m, t), p( m, m, 3), a® m, 1 1
fc(fc*4>ffi}
12.5 Symmetric Equation* for a Line 543
■ In Exercises 61 0, a line is given. To which of the coordinate axes, if any, is
it parallel?
10. fa y, z) : x = 3, y = 4}
■ In Exercises 1 1— IS, a Hue is given in terms of parametric equations. Repre
sent the line using symmetric equations.
11. {(,*, y,z): x = 1 f 3k, r/ = 2 — 2k, z = 3 — 4Ar„ fc is real}
12. {{x, y,z):x =  1  k, y = 2 + 4k, z = 5  k, k is real}
13. ((x. {/, z) : .t = A\ y s 2k t * = 3fc, fc is real}
14. {(x, ^ z) : x = 3  4k y y = 2fc, 2 = 1  k, k is real)
15. {x, y. z) : x = 2 + ft(5  2), y = 3 + Jk{5  3), 2 9 2 + fc(l  2),
k is real}
■ In Exercises 1620, a plane and a line are given. Find their point of
intersection.
lfi. ((x, i Jt z) : 3x + 4y + 55 = 80}, {(*, I/, jj : =  « )
17. {{x, j/, 5) : 2x 1 y  10 = 0} T {(x T r/, s) ; * = 5 and 3 = 3}
18. {{x, ,j,z):xy + z = 0}, [{x, y, z) ; ^1 = ±^A = i^.}
16. {(x, y, z) : 2x  y + z + 5 = 0},
20. {{x, j/,s) : x + «/ + 3 + 1 = 0), {{x, $f. 3) : % = y = z)
21. Prove Theorem 12.9,
22. Prove Theorem 12,10.
23. Prove Theorem 12.11.
24. Prove Theorem 12.12.
25. Prove Theorem 12.13.
544 Coordinates in Space Chapter 12
CHAPTER SUMMARY
We began this chapter by defining an xj/jbCOORDINATE SYSTEM.
Stalling with a pair of perpendicular lines called the x and the yaxcs, we
know that there is one and only one line through their intersection that is
perpendicular to both of them* We call this line the zAXIS, The one and
only point that lies on all three of the coordinate axes is called the ORIGIN'.
There are three coordinate axes and three coordinate planes. Each axis lies
in two of the coordinate planes and is perpendicular to the oilier one. All
lines perpendicular to a coordinate plane are parallel to one of the coor
dinate axes. AH lines parallel to a coordinate axis are perpendicular to one
and the same coordinate plane. If a plane is parallel to a coordinate plane,
then it is perpendicular to a coordinate axis and to all lines parallel to that
axis.
In our work with coordinates in space we stated and proved the
DISTANCE FORMULA THEOREM and slated a theorem regarding
parametric equations for a line, These theorems are natural extensions of
some theorems for the twodimensional case in Chapter 11, The Distance
Formula and parametric equations for a line are useful tools in solving prob
lems involving solid geometry. Linear equations may be used to represent
planes, and combinations of them in what we call symmetric form may be
used to represent lines.
REVIEW EXERCISES
In Exercises 18, eight points are given as follows: A(0, 0, 0), 13(3, 0, 0),
C{3, 4, 0), D(Q, 4, 0), £(0, 0, 2), F(3, 0, 2), G(3, 4, 2), H(0, 4, 2).
1. Draw a graph of the rectangular solid with vertices A. B, C, D, E. F,
G,H.
2. Prove that lABF^ L DBF — / CBF =z LEFC.
3. Find the measure of dihedral angle IIEAF,
4. Prove that AG = EC ss W =s ¥D.
5. Find the distance from point A to plane CDH.
& Find the distance from point A to plane FOIL
7. Find the distance hetwecQ plane ABC and plane EFC2.
8. l^et I. J, K be the midpoints of EE, ItfZ, DC ( respectively. Find the dis
tance between the planes JJK and ECC.
In Exercises 913, find the distance between the two given points,
9. (Xi, t flt %) and (jfifc t/ 2 , z*). 12. (7, 8, 0) and (10, 12 t 0).
10, (a, b, c) and (d, e, /). 13. (3, 4, 5} and (8, 9, 10).
11. (p, q, r) and (p t q, t).
Review Extrclsts 545
In Exercises 1422, there are two given points: A{1, 2, 3) and #(4. 5, 6)*
14. Express AB using parametric equations and setbuilder notation.
15. Express AB using parametric equations and setbuilder notation.
16. Express BA using parametric equations and setbuilder notation,
17. Express opp AH using parametric equations and setbuilder notation.
18. Express opp BA using parametric equations and setbuilder notation.
10. Find C on AB if AC = 10  AB.
20. Kind C on opp AS it AC = 10 * AB.
21. Find the bisection points of AE.
22. Find the two points winch divide AB internally and externally in the
ratio ,.
23. Given A(5, 1.1). fl{3, 1, 0), C<4. 3, 2). JJ(6 t 3 T 1}, prove that quad
rilateral ABCD is a parallelogram.
24. Given A(2, 4, 1), B(l, 2, 2), C(5> 0, 2), pro%e that AABCis a right
triangle,
25. Find three noncollincar points on the plane given by the equation
3x + Ay  z + 15 = 0.
26. Find three distinct points on the line given by the symmetric equation
x—2 _ y4 _ z+3
3 ~ 2 5 '
27. Write an equation of the plane that is perpendicular at (2, 4, — 3) to the
line of Exercise 26.
28. Find the x, y, and sintercepts of the plane a where
a = {(jr, tj t s) : 3x  7y  fa + 5 = 0}.
29. Express the line / where
I = {(*, sj, z) : x = 1 + 2*, y = 1  3fc, z = k,k is reu]},
using symmetric equations but no parameter.
3(1. Consider the line I and the plane or perpendicular to / at C>(0, 0, 0), given
as follows:
Find two distinct points P and Q on J such that « is the perpendicular
bisecting plane of T$,
/
\
4
Chapter
Erick ttartmoimf Magnum Photot
Circles
and Spheres
13.1 INTRODUCTION
An the title suggests, this chapter is concerned with properties of
circles and spheres, some of which you may already have studied in
your earlier work in mathematics.
The first part of the chapter deals with the intersection proj)erties
of a circle and a line in the plane of the circle and the intersection
properties of a sphere and a plane.
The second part of the chapter is about the degree measure of arcs
of a circle and properties of certain angles in relation to arcs, secants,
tangents, and chords. We also consider properties of lengths of secant
segments, tangentsegments, and chords.
13.2 CIRCLES AND SPHERES: BASIC DEFINITIONS
Up to now, in our formal geometry, we have not discussed "curved
figures/' that is, figures not made up of segments, rays, or lines* Perhaps
the simplest of these figures are the circle, the sphere, and portions of
a circle, called arc*;. We begin with some formal definitions.
548 Circles and Spheres
Chapter 1 3
Definition 13.1 (See Figure 131.) Let r be a positive num
ber and let O be a point in a given plane. The set of all points
Fin the given plane such that OP — r is called a circle. The
given point O is called the center of the circle, the given
number r is called the radius of the circle, and the number
2r is called the diameter of the circle.
A circle is a curve and encloses a portion of a plane. As an illustra
tion, consider a circular disk. The edge of the disk is what we have in
mind when we think of a circle. The edge of the disk together with its
interior points is what we have in mind when we think of a circular
region. Wc will have more to say about a circular region (that is, a circle
and its interior) in Chapter 14,
Question: Is the center of a circle a point of the circle? Explain,
Circle with center O
and radius OF = r
^V
Sphere with center
and radius, OP = r
Figure 131
Definition 13.2 (See Fij^ire 131.) Let r be a positive Dum
ber and lei () be a point in space. The set of all points P in
space such that OP = r is called a sphere. The given point O
is called the center of the sphere, the given number r is
called the radius of the sphere, and the number 2r is called
the diameter of the sphere.
A sphere is a surface and encloses a portion of space. As an illus
tration, consider a ball. The surface of the ball is what we have in mind
when we think of a sphere. The surface of the ball together with its
interior points is what we have in mind when we think of a spherical
region. We will have more to say atout a spherical region (that is, a
sphere and its interior) in Chapter 15,
13.2 Circles and Spheres: Banc Definitions 549
Definition 13.3 Two or more coplanar circles, or two or
more spheres, with the same center are .said to be concentric.
In Figure 132, Q is the common center of three concentric circles
with radii (plural of radius) ri, r 2l and r 3 „ respectively.
Figure 132
Let O be a point in a given plane a. Then we can choose an xycoor
dinatc system in a such that the origin is at O. Let Cbe a circle in a with
center at O and radius r. Let P(x, y) be any point on C as shown in
Figure 133.
Pto, y)
Simon 133
Then, by the Distance Formula, OP = r = >/(x  0) + (y  G) 2 .
Therefore r = \/^~+ y 2 and x 2 + y = r 2 .
Converse ly, if F(x, y) is any point such that x 2 + y* = f*, then
y/( x  0) 2 + (y  0) 2 = r T OF = r, and F is a point of the circle C.
Thus, in a given xt/plane, the circle C with center at the origin and
with radius r is given by C = {{x, y) ; x 2 + y 2 = r 2 }.
We have proved the following theorem.
550 Circles and Spheres.
Chapter 13
THEOREM 13.1 Let an xyphme be given and let O be the origin
and let r be a positive number. Let C be the circle in the syplane
with center O and radius f. Then
C = {(*, y) : ** + r/2 = r*}.
Definition 13.4 A chord of a circle or a sphere is a segment
whose endpoints are points of the circle or sphere. A secant
of a circle or sphere is a line containing a chord of the circle
or sphere A diameter of a circle or sphere is a chord contain
ing the center of the circle or sphere. A radius of a circle or
sphere is a segment with one endpoint at the center and the
other endpoint on the circle or sphere.
In Figure 134, C is a circle with center P, and S is a sphere with
cent er P. For the circle and the sphere, AB is a chord, 46 is a secant,
ED is a diameter (and also a chord), and PQ is a radius. The endpoint of
a radius that is on a circle or a sphere (such as point Q in Figure 134)
is often referred to as the outer end of that radius. Is ED in Figure 134
a secant? It follows from Definition 13,4 that a secant is a line that in
tersects a circle (or sphere) in two distinct points.
*<*
Figure 13*
Note that, in connection with a circle or a sphere, the word "radius"
is used in two different ways and the word "diameter" is used in two
different ways: (1) each word 1$ used to mean a certain segment and
(2) each word is used to mean the positive number that is the length
of a segment. Hus should not be confusing because the context in
which the word is used should make it easy for you to decide which
meaning Ls intended. For example, if we speak of a radius or a diameter
of a circle or of a sphere, we mean a segment. If we speak of the radius
13.2 Circles and Spheres: Basic Definitions 551
or the diameter, we in can the number that is the length of a segment.
Thus a circle has infinitely many different radii if radius is interpreted
to mean a segment; it has just one radius if radius means the length of a
segment.
Just as we speak of congruent angles, or congruent segments, or
congruent triangles, we often speak about congruent circles or con
gruent spheres. How would you define congruent circles or congruent
spheres? Does your definition agree with the following one?
Definition 13,$ Two circles (distinct or not) are congruent
if their radii arc equal. Two spheres (distinct or not) are con
gruent if their radii are equal.
Using Definition 13.5, it is not hard to prove that congruence for
circles (or spheres) is an equivalence relation; that is, it is reflexive, sym
metric, and transitive. Thus if A, B, C are any three circles (or spheres),
it is true that
I A=A.
2. IfA^B, thcnBsA.
3. Tf A == B and B s C, then AsC.
Reflexive Property
Symmetric Property
Transitive Property
EXERCISES 13.2
In Exercises 1 10, refer to the circle with center F shown in Figure 135.
Assume that all the points named in the figure are where they appear to l>e
in the plane of the circle. Copy and replace the question marks with words
or symbols that best name or describe the indicated parts.
1. TJE is called a [7] of the circle.
2. FE is called a [T] of the circle.
3. AH is called a [Tj of the circle.
AB could also be culled a (T]
of the circle.
4* FG is called a (7] of the circle.
S. FG is called a (7] of the circle.
& DE is called a [7] of the circle.
7. C is the outer end of the radius [TJ*
8. A is the 7J of the radius 7].
9. The points named in the figure thai arc points of the circle (that is, on
the circle) are 7],
10. The points named in the figure that arc not points of the circle are [?]■
li^m. U.
552 Circles and Spheres
Chapter 13
Li Exercises 1115, refer to the Sphere with center Q shown in Figure 136.
Copy and replace the question murks with words or symbols that best name
or describe the indicated parts.
Figure 139
11. QR is a [Tj of the sphere.
12. If S, Q, T are collincar, then ST is a [TJ of the sphere,
13. KS is a \?} of the sphere,
14. RT h a [?] of the sphere.
15. Points T are outer ends of given radii.
16. Prove that if two circles are congruent, then a diameter of one is con
gruent to a diameter of the other.
17. Prove that congruence of circles is an equivalence relation, (You will
need to use Definition 13.5 and tbe theorem that congruence of seg
ments (radii) is an equivalence relation.)
In Exercises 1825, refer to the circle with center at the origin of Lhc xy
conrdinatc system shown in Figure 137, P{x t o) is a point on the circle.
Figure 13*7
!
<
^y
j*
P(
"K
/ 
V *
{
\*
q
6 14 .3.31 t
12 3 4
1 6 '
\

1
/
1
1
13.2 Circles and Spheres: Basic Definitions 553
18. Use the Distance Formula and express the distance between O and P
in terms of x and y.
19. Write an equation like that in Theorem 13. 1 of the circle with center
at the origin and radius 5.
20. Write the coordinates of the points on the x and yaxes that arc on the
circle of Exercise 19. Check your answers by substituting the coordi
nates in the equation of the circle.
2L Is A = (3, i) u point of the circle of Exercise 19?
22. Is B = ( — 4, 3) a point of the circle of Exercise 19?
23. Is K = (l t 2%/^) a point of the circle of Exercise 19?
24. Is R = (—2, 4.5) a point of the circle of Exercise 19?
25. Write the coordinates of two more points (different from those in Ex
erases 2024) that lie on the circle of Exercise 19.
26. Given the ,t [/coordinate system shown in the figure with A = (1,2),
P b (x, y) y and AP — 5, use the Distance Formula and express the dis
tance between A and Fin terms of x and y,
'
'>'
,
.
4 «*.*j
4
*\
4(1.2)
x *
21
12 3 4
s a
■
i
27. In Exercise 26, write an equation like that in Theorem 13.1 of the circle
with ecntcr at A and radius AP = 5. Is the point S = (4, 6) a point of
this circle? Write the coordinates of at least three more points that are
on this circle.
25. Which of the following are points of C as {{%, y) : x 2 + y 2 = 100}?
(a) (a 10) (b) (6,8) (c) (4,9) (d) (2^ 2\/I5)
29. Whet is the nidi us of the circle of Exercise 28?
30. Find five more points of C in Exercise 28.
Exercises 3134 concern the set C, where C  {(x, y) : i* + y 2 = 16).
31. Is C a circle? Why?
32. Find x if {x, 3) is a point of C. (There are two possible values for a ;
33. Find y if (4. y) is a point of C.
34. Can you find x so that (x, 5) is a point of C? Explain.
554 Circles and Spheres Chapter 13
■ Exercises 3537 concern the circle C» where C = {(x, y) : x 2 + y z = 36).
35. What restrictions on x and t/ would give only the part of the circle in
quadrant 1?
36. What part of the circle would you be considering under the restriction
x 2 + y 2 = 36 and x < 0?
37. What restrictions ou x and y would give the intersection of C and quad
rant 111?
38. challenge problem. Let an xi/plane with A = {h, k), F = (x t y),
AP = r > be given. Write an equation in terms of x, y, h f k t r for the
circle with center al A and radius AP = r.
13,3 TANGENT LINES
If you look at a drawing of a circle in a given plane, it is easy to see
that the circle separates the points of the plane not on the circle into
two sets. One of these sets consists of those points of the given plane
that are "inside" the circle and the other set consists of those points
of the given plane that are "outside" the circle.
Definition 13.6 (See Figure 138.) Let a circle with center
O and radius r in plane a he given. The interior of the circle
is the set of all points P in plane a such that OP < r. The
exterior of the circle is the set of all points P in plane « such
that OP > r.
Home 13S
It is clear from Definitions 13,1 and 13.6 that if F is anv point in
the plane of a circle with center O and radius r, then P is on the circle
{OP = r), or P is in the interior of the circle (OF <[ r), or P is in the ex
13.3 iMpii Lines 555
terior of the circle {OP > r). Wc sometimes say "Pis inside the circle'
or "P is outside the circle*' when we mean "P is in the interior of the
cirde" or "Pis in the exterior of the circle. 1 ' respectively.
Figure J 39 shows an ^coordinate system whose origin is the
center of a given circle C with radius r. The figure also includes expres
sions for C, its interior and its exterior, in terms of coordinates using
set Wilder notation.
Inferior of C =• {(*; Jf)
F.xlerk* at C = {(*»jr)
«» + 9* = »*)
•* + V* > **)
Figure 139
Definition 13.7 If a line in the plane of a circle intersects
the circle in exactly one point, the line is called a tangent to
the circle and the point is called the point of tangeney,or
die point of contact. Wc say that the line and the circle arc
tangent at this point. J f a segment or a ray intersects a circle
and if the line that contains that segment or ray is tangent to
the circle, then the segment or ray is said to lx* tangent to
the circle.
In Figure 1310, if Ms a line in the plane of circle C and if the in
tersection of I and C Is just the one point P, then I is tangent to C at P.
Figure 1310
5 56 Circles and Spheres
Chapter 13
Given a circle and a line in the plane of the circle, what are the possi
bilities with regard to the intersection of the line and the circle? Figure
131 1 suggests that there arc only these three possibilities: the inter
section of the line and the circle may be the empty set, or it may be a
set consisting of exactly one point, or it may be a set consisting of
exactly two points.
cnl=\P\ t
blgwo 1311
That these arc the only three possibilities can be verified in the fol
lowing way. Suppose that wc are given a circle C with center in plane
a and that P is a point in plane a. Then P is outside the circle, as shown
in Figure 131 la, or P is on the circle, as shown in (b) t or P is inside
the circle as shown in (e).
If Pis outside the circle, we shall show that the unique line I in plane
cr such that OP _^ I at P does not intersect the circle. If P is on the cir
cle, wc shall show that the unique line / in plane a such that OP ± I at
F intersects the circle in exactly one point (hence / is tangent lo C at P).
If f is inside the circle, then either P = O or P=£CK1( P / 0, we
will show that the unique line I in plane a such that OP I I at P inter
sects die circle in exactly two points which are equidistant from P,
Finally, if P = O, we will show that any line I which contains P inter
sects the circle in exactly two points. We are now ready for the follow
ing theorem.
33.3 Tangent Lines 557
THEOREM 13.2 Given a line I and a circle C in the same plane,
let O be the center of the circle and let P l>e the foot of the per
pendicular from O to line I,
L Every point of 2 is outside C if and only if P is outside C
2. I is tangent to C if and only if P is on C
3. I is a secant of C if and only if P is inside C.
Proof: Let r be the radius of C and let OP = a. We select an ^coor
dinate system in the plane of C and I with the origin at O s with the
yaxis parallel to I and with P on the nomiegative arasis. Then
and
P=(a,0)>
Z= {(*,*/) :* = a}.
Proo/ 0/ i; Suppose we are given that ? is outside C as shown in Fig
ure 13*12; then a > r]> 0. Why? It follows that a 2 > r 2 and hence
Q 2 + y 2 > **• Therefore all points (fl* y) are outside C, Since
t{{x i y):x=a} = {(a, y) : y is real},
it follows that all points of J ate outside C.
figure 1312
Now suppose that I is outside C; then every point of I, including P %
is outside C and the proof of (1) is complete.
558 Circles and Sphert*
::>spter ".3
Proof of 2: Suppose wc are given I hat F is on C as shown in Figure
1313; then a = r. Why? The intersection of C and I is
Ifa y) : x* f i/ = r*} fl {(*, y):x = u} = {(a. y):y* = Q}>
c
Ij
Pte.D)
FlRUIT 1313
Since zero is the only number whose square is zero, it follows that
y = 0. Therefore the only point of intersection of I and C is P{m 0).
Thereforc I is tangent to C at P. Why? This proves that I is tangent to
C if P is on C,
Now suppose it is given that I is tangent to C. Then J and C have
exactly one point in common. This means that I H Cis a set consisting
of exactly one point. But
C = {{*, y) : x* + f = i*},
and
IDC= {{x, \j) : x* 4 J/ 2 = r 2 and* = a}
{fey):* 4 E/ 2 = r2}.
If (a, tj) with t/ ^ Q is a point of Z fl C, then (a, y) is also a point of
I H C and there are two distinct points in I D C. Since there is only
one point in / fl C, it follows that y = and a z = #, Since a > 3 it
follows that a m f and the one and only point in I H C is the point
HA 0)* This proves that / is tangent to C only if P is on C,
13.3 Tangent Un« 559
Proof of 3: Suppose w© are given that P is inside C as shown in Figure
1314a or b. Recall that O is the center of the given circle C Then
P = («, 0) where < a < r>
an<
I n G = {(*, y) : x = a and * 2 + t/ 2 = r 2 }
Since < a < r, il follows that
r*  o 2 > 0, y/fid* ^  Vr^fl*
and the points (a, "v 71,2 — ° 2 ) an ° 1 (*>■ — V 4 — ° 2 ) are distinct points.
This proves that / is a secant of C if P is inside C.
Suppose now that I is a secant of C; then / intersects C in two dis
tinct points which we have just shown to be (a. yV 2 — a 2 ) and
{a> — y/f& — « a ). This means that r 2 > « a . (Why is it that we cannot
have r a = a 2 or r 2 < a 2 ?} Since r > and « > 0, it follows that r > a
and hence that OP <C ft Therefore F(a, 0) is inside C and the proof is
complete.
Now that we have proved Theorem 13.2 we proceed to state our
first basic theorems on tangents and chords. To prove some of these
theorems wc need refer only to Theorem 13.2 and see which of parts
(1), (2), or (3) apply.
560 Circles and Spheres
Chapter 13
THEOREM 13.3 Given a circle and a line in the same plane, if
tlie line is tangent to the circle, then it is perpendicular to the radius
whose outer end is the point of tangency.
Proof: Let C be the given circle with center O, let line I be tangent
to C at R, and let P be the foot of the perpendicular from O to /. it fol*
lows from part 2 of Theorem 13.2 that P is on C. (See Figure 1315.)
The figure shows R and P to be
distinct points. We shall prove
that they are the same point.
If Pi=R, then OR is the
hypotenuse of right triangle
AOPR and OR > OP, But R
and P are both on C. Therefore
OR = OP. Since this is a con
tradiction, it follows that P = R.
Since I ±OP at P t it follows
that I ± OR at R, and the proof
is complete.
Figure 1315
Our next theorem is the converse of Theorem 13.3.
THEOREM 13.4 Given a circle and a line in the same plane, if
the line is perpendicular to a radius at its outer end, then the line
is a tangent to the circle.
Proof: Assigned as an exercise.
THEOREM 13.5 A diameter of a circle bisects a chord of the
circle other than a diameter if and only if it is perpendicular to the
chord.
Proof: Let a circle C with radius r, center O, diameter AJ3, and chord
RQ be given. Wc must prove two things.
Ll£AB±KQ t then A75 bisects ftg.
2. If AB bisects RQ, then A~B ± RQ.
Proof of 1 : Choose an ^coordinate system in the plane of the given
circlc_C such that the origin is at the center 0,A = {  r, 0), B = (n 0),
and RQ is perpendicular to KB at P(a, 0), where < a < r as shown
in Figure 1316, Then AB is a diameter of the circle. "Why?
13.3 Tangent Lirm 561
Figure 1316
Suppose /? and {) are named so that R is in the first quadrant. Then
(see the proof of part 3 of Theorem 13,2) R = (a, y/& — a 2 )
Q = (a, — ^/r* — a 1 }. Therefore
and
FR = JVf«fl 2 0 = V* " o*
Pp = p  ( Vf2  a*)j = v^^ 1 ^
Therefore Fit = FQ and AB bisects ff^ at P. This proves that if a di
ameter is perpendicular to a chord, then it bisects the chord.
Proof of 2; Choose an arycoordinate system in the plane of the given
circle C such that the origin O is the center of the circle, A = ( — r, 0),
B = (r, 0), and the chord EQ (not a diameter) intersects AB at P{a, 0)
with < a < r, (See Figure 1317.)
Figure 1317
562 Circles and Spheres Chapter 13
Suppose that P is the midpoint of RQ. Then If R = (x^ ^) and
Q — (** Sfe)* *■ nave
Pfl = Pp,
(*!  a)> + kl! = (* 2  a ) 2 + y 2 2 , (Why?)
*i 2  2*ia + 0* + y,a = Xi 2  2x& + a 2 + y#.
But xi 2 + tji 2 = f 2  *a a + yz 2  (Why?)
Then — 2x\a a — 2^ 2 a> *i = **. RQ is a vertical line, and RQ I
AB, This proves that if a diameter bisects a chord that is not a diameter,
then the diameter and the chord are perpendicular. This completes
the proof of Theorem 13.5,
THEOREM 13,6 In the plane of a circle, the perpendicular bi
sector of a chord contains the center of the circle.
Proof: In the proof of part 1 of Theorem 13.5, AB is the perpendic
ular bisector of chord BQ in the plane of the given circle by the defi
nition of the perpendicular bisector of a segment in a plane. Since An
contains O, the center of the given circle. Theorem 13.6 is proved*
TIJEOREM 13.7 Let a circle C and a line I in the plane of the
circle be given. If I intersects the interior of C, then I intersects C
in exactly two distinct points.
Proof. Theorem 13.7 follows from part 3 of Theorem 13,2. The de
tails of the proof are assigned as an exercise.
THEOREM 13,8 Chords of congruent circles are congruent if
and only if they are equidistant from the centers of the circles.
Proof: Let C" and C be the given congruent circles with centers P
and F and radii f and f, respectively. Then r = f. Why? Let AB and
A'B' be chords of the given circles C and C, respectively. Suppose,
first that the distance from Ali to the center of circle C is zero, that is.
AB is a diameter of C, and suppose that AU ^= A'B'. Then
AB = A'B' = 2r = If
and ~ATE is a diameter of circle C Therefore AB and A'B' arc equi
distant from P andF (th e dista nces being in this case). Conversely., if
the distances of AB and A'B' from P and F are 0, then AB and /PF
are diameters of C and C. It follows that AB = 2r = 2/ = A'B 1 ;
hence AB === A'B',
13.3 Tangent Lines
Now suppose that AB is not a diameter of C. If AB s A'B\ it fol
lows that A'B' is not a diameter of C, Whv? Let F be the foot of the
perpendicular from P to AB and let F be the foot of the perpendicular
from F to A'B' as shown in Figure 1318. Then by Theorem 13 .5,
A** = l 2 AB and A'F = JA'B'.
5*3
Figure 1 3 IS
By hypothesis,
"Therefore
We have
AB = A'B'.
AF = A'F.
AP = A'F.
(Why?)
Since AAPF and AA'FF are right triangles, it follows that AAPF
ss AA'FF by the HypotenuseLeg Theorem. Therefore PF = FF
and hence AB and A'fr are equidistant from F and F,
Conversely, if KB and A'B' are equidistant from P and F, that is,
if
Pf?=FF#0 (
ft follows that AAPFss AA'FF by die HypotenuseLeg Theorem,
(Show that AAPF & AA'FF if PF = FF.) Therefore AF = A'F.
But by Theorem 13.5,
AF =
and A'F = ^A'F.
AB = A'B' and ^EssA'/?'.
This completes the proof of Theorem 13.8,
It should be noted that the congruent circles of Theorem 13.8 could
be the same circle, in which case the theorem still holds.
564 arete* and Spheres
Chapter 13
Figure 1319 shows two different examples of two circles tangent
to the same line at the same point In Figure 1319& the centers A and
A' of the two circles are on the same side of the tangent line I and the
circles are said to be internally Umgent, In Figure 1 3 1 9b the centers
B and B' of the two circles are on opposite sides of the tangent line n
and the circles are said to be externally tangent.
[Iml.ujIIj, Rutgeal 
m
Figure 1319
Our formal definition follows.
Definition 13.8 l\vo circles are tangent if and only if they
are coplanar and tangent to the same line at the same point
If the centers of the tangent circles are on the same side of the
tangent line, the circles arc said to be internally tangent. If
their centers arc on opposite sides of the tangent line, the
circles are said to be externally tangent.
EXERCISES 13.3
Exercises 110 refer to the circle C •= {(*, tf) : x 3 ■§■ tf = 64}. In each ex
ercise, the coordinates of a point arc given. Tell whether the point is en the
circle, in the interior of the circle, or in the exterior of the circle.
h (0, 8)
4. (4,7)
7. (0, 8)
10. (6. 2\/7)
2, (3, 5)
5. (4v^,4)
8. (8, 1)
3. (7,3)
G, (4V3, 4)
9. (6, 6)
11. Find the endpoints of two distinct diameters of the circle
C = {(*» y) : ** + y* = ea } .
12, Use setbuilder notation to express the set of points in the exterior £
of the circle C = {(*, tj) : x' + tf = 64}.
13,3 Tangent Un*s 565
13. Use set builder notation to express the set of points in the interior I of
the circle C ■ {(x, y) : x 2 + y 2 = 64}.
14. Prove that the center of a circle is in the interior of the circle.
Exercises 1520 refer to the circle C" with center tit (0, 0) and passing through
the point ?( 5, 12).
15. Find the radius of the circle.
16. Use setbuilder notation to express the set of points on the circle.
17* Find eight distinct points that are on the circle
18. Find two distinct points that are in the exterior of the circle.
19. Find two distinct points that are in the interior of the circle.
20. If / is the tangent line to C at P(— 5, 12). find an equation for L (Hint;
I — UP at P, Find the slope of I and use the Point Slope Form of an
equation,)
21. Ut the circle C = {(*, y) : x 2 f y* = 36} and the line
I = {(x, y) : x = 3} be given,
(a) Does I intersect C?
(b) If / intersects C, is I a tangent line or a secant line?
(c) If I intersects C, find the coordinates of the poiutjs) of intersection.
22. Let the circle C = {(x, y) : x* + y 2 = 25} and the lines
Ei = {(*.y) : x = 5}, h= {(*,$f) : y = 2*}, h = {fry) : * = 7)
he given.
(a) Which of the lines intersect the circle?
(b) Which line is tangent to the circle? What are the coordinates of the
point of Langency?
(c) Which line is a secant? What are the coordinates of the points of
intersection of the secant line and the circle?
23. Prove Theorem 13.4.
24. Prove Theorem 13.7,
25. Copy and complete: A tangent to a circle is JT] to the radius drawn to
the point of contact.
26. Copy and complete: If a diameter is perpendicular to a chord, then it
[?] the chord.
27. Copy and complete: If a diameter bisects ft chord other than a diameter,
then it Is 3.
28. Copy and complete: In the plane of a circle, the perpendicular bisector
of a chord contains the [T].
29. In a circle with r ad his 13 in., how long is a chord 5 in. from the center
of the circle?
30. In a circle with diameter 12 cm., how long is a chord 4 cm, from the
center of the circle?
3L Find the radius of a circle if a chord 8 in. long is 3 in. from the center of
the circle.
566 Circles and Spheres
Chapter 13
32, How far from the center of a circle with a radius equal to 25 is a chord
whose length is 30?
33. In the figure. FQ is parallel to I which is tangent to the circle at F. The
center of the circle is R and FQ bisects RF at Jtf. If FQ = IS, find BP,
[Hint: Let RF = 2x. Then RP = 2x and RM = x.)
w
Jl
34. If AH is a diameter of a circle and if lines t t and i* arc tangent to the
circle at A and B t respectively, prove h \\ £&.
35. Prove that the line containing the centers of two tangent circles con
tains the point of tangeucy. (See Figure 1319.)
36. The figure below at left shows two concentric circles, AB is a chord of
die larger circle and is tangent to the smaller circle at AC Prove that M
is the midpoint of A~B t
37. In the figure above at right, T is a point in the exterior of the circle with
center P. T\vo distinct tangents are drawn from T to the circle with
points of contact A and B. Prove that TF bisects LATB and that
AT = BT.
3S. In Exercise 37. if the radius of the circle is 9 and FT = 15, find AT.
39. challenge problem. In Exercise 37 if the radius of the circle is 9 and
FT = 15, find AH.
13.4 Tangent Plana* 567
40. CHALLENGE PROBLEM. Show thilt the lillC I = {(x. If) I 3x 4 4(/ = 25}
i8 tangent to Uic circle C = {(*, tf) : ** + y 2 = 25) and find the co
ordinates of the point of tangency.
41. challenge problem. Prove that no circle contains three col linear
points. {Hint: Use Theorem 13.6.)
13.4 TANGENT PLANES
In Section 13.3 we studied relations between lines and circles in a
plane. In this section we study relations between planes and spheres in
space. There is a close analogy between the definitions and between
the theorems of the two sections.
Onr first theorem of this section is analogous to Theorem 13.1, In
Chapter 12 you learned how to find the distance between two points in
space by introducing a threedimensional coordinate system (called an
$j/£coordinatc system). We shall use this Distance Formula to develop
an equation for a sphere in an xy£coordtaate system.
Let O be a point and r a positive numl ier. Let S be the sphere with
center O and radius r. Suppose an ^coordinate system has been set
up with O as the origin. (See Figure 1320.) Then Pfc tj, %) is a point of
S if and only if
OF = s/{x  0)2 + (y  0)8 + (Z  0)' = r s
V* 2 + tf 3 + « a = r,
* a + y 2 4 z 2 = r 2 .
P(x,y,»ji
Figure 1330
We have proved the following theorem.
568 Circles and Spheres
Chapter 13
THEOREM 13.9 Let be a point, r a positive number, and S the
sphere with center O and radius r. Given an jcy.scoordmate system
with origin O, S = {(ar, y, z) ; x 2 + y 2 + s 2 = r 2 }.
Definition 13M (See Figure 1321.) Let a sphere with
center and radius r be given. The interior of the sphere is
the set of all points P in space such that OF << r. The exterior
of the sphere is the set of all points P in space such that
OP>r.
e
J:.,'Tk,r
*\. Exterior
<>
^
Figure 1321
In view of Definition 139 and Theorem 13.9. if S is a sphere with
radius r and center at the origin of an 3fyscoordinate svstem, then
S = {(*, if, z) : X 2 + tf + *2 = r 2 }
Z= {(*,y f *):a* + if + ^<ia}
and
where / is the interior of the sphere and £ is its exterior.
Before reading Definition 13.10, try to form your own definition
of a plane tangent to a sphere. See Figure 1322.
Figure 1323
13.4 Tangent Plants 569
Definition 13.10 If a plane intersects a sphere in exactly
one point, the plane is called a tangent plane to the sphere.
The point is called the point of tangent \\ or the point of
contact, and we say that the plane and the sphere are tangent
at this point.
You have seen that there are three possibilities with regard to the
intersection of a line and a circle in the same plane. Figure 1323
suggests three possibilities with regard to the intersection of a plane
and a sphere. For each part of Figure 1323. S is a sphere with center O
and P is the foot of the perpendicular from O to plane a. Figure 1 323a
suggests that if P is in the exterior of the sphere, then all of plane a is in
die exterior of the sphere and S n n = . Figure 1323b suggests that
if jP is on the sphere, then a is tangent to the sphere at P and Sfla =
{P}. Figure l323c suggests that if P is in the interior of the sphere,
then the intersection of the sphere and plane ot is a circle C with center
P, that is, S n a = C.
a
(a)
(b)
(O
Figure 1333
The following theorem concerning the intersection of a plane and
a sphere is analogous to Theorem 13.2 concerning the intersection of a
line and a circle in a plane. It can be proved using a threedin ie us it mal
coordinate system in much the same way that Theorem 13.2 was
proved using a twodimensional coordinate system. You will be asked
to write a proof of Theorem 13.10 in the Exercises.
THEOREM 13.10 Given a sphere S with center O and a plane a
which docs not contain O, let P be die foot of the perpendicular
from O to a.
L Every point of a is in the exterior of S if and only if P is in the
exterior of S.
2* a is tangent to S if and only if P is on $♦
3. a intersects S in a circle with center P if and only if P is in the
interior of S,
570 Circles and Spheral Chapter 13
1HEOREM 13. J J Let a sphere S with center O and radius r and
a plane a be given. If the intersection of S with a contains the center
O of the sphere, then the intersection is a circle whose center and
radius are the same as those of the sphere.
Proof: Let a sphere S with center O and radius r be given as shown in
Figure 1324. Let a be a plane that contains O and intersect*. S. Since S
is the set of all points of space whose distance from O is r, the inter
section of S and a is the set of all points P of a such that OP = r. By
definition, this set is the circle in plane a with center O and radius r.
Thus the circle has the same center and radius as the sphere, and the
proof is complete*
Figure 1324
Definition 13.11 A circle that is the intersection of a sphere
with a plane through the center of the sphere is called a great
circle of tile sphere.
The next two theorems can be easily proved using Theorem 13, 1 1.
THEOREM 13.12 The perpendicular from the center of a sphere
to a chord of the sphere bisects the chord.
Proof: The endpoints of the given chord and the center O of the given
sphere determine a plane O. The intersection of plane a with the sphere
is a great circle: with center O and having the same chord as the given
chord It follows from Theorem 13.5 that the perpendicula: Irom ( Mo
the given chord bisects the chord.
THEOREM 13.13 The segment joining the center of a sphere to
the midpoint of a chord of the sphere is perpendicular to the chord.
Proof: Assigned as an exercise.
13.4 Tangent Planes 571
Our last theorem of this section is analogous to Theorems 13.3
and 13.4.
THEOREM 13. 14 A plane is tangent to a sphere if and only if it
is perpendicular to a radius of the sphere at its outer end.
Proof: Let ft sphere S with center O and radius OF be given. Let «
tie the given plane. There are two parts to the proof.
1. If UP A a at P, then a is tangent to S at P.
2. If a Is tangent to 5 at P y then OP _ a.
Proof of h We are given that OP J a al P as shown in Figure 1325.
Let R be any point of a different from P; then OP X M (Why?) and
AOPR is a right triangle with the right angle at P. Therefore
OP < OR. Why? Therefore ii is a point in the exterior of S. Why? It
follows that P is the only point of a that belongs to boil a and S; hence
0! is tangent to S at P.
Figure 1323
Proof of 2: We are given that a is tangent to S at P. We shall use an
indirect proof to show that OP ± a, Suppose, contrary to what we
want to prove, that OP is not perpendicular to a. Let Q be the foot of
the unique perpendicular from O to a. Then OQ < OF. Why? There
fore Q is in the interior of S, Why? It follows from part 3 of Theorem
13.10 that a intersects S in a circle. But this contradicts the hypothesis
that a is tangent to S; that is, the intersection of a and S is exactly one
point. Therefore our supposition that UP is not perpendicular to a is
incorrect and we conclude that OP _L a. This completes the proof of
Theorem 13.14,
572 Circles and Spheres Chapter 13
EXERCISES 13.4
L I*et a sphere with radius 10 in, lie given, A plane 6 in. from the center
of the sphere intersects the sphere in a circle. Find the radius of the
circle.
2. Given a sphere S with center O, let a and B be two planes equidistant
from O and such that a intersects S in a circle C and /? intersects S in a
circle C. Prove that C is congruent to C. {Hint; Let F be the foot of the
perpendicular from O to o and let F be the foot of the perpendicular
from O to jS. Let G be a point on C and C" be a point on C. Prove that
&Omm &OFC.)
3. In Exercise 2, is it necessary for « and /? to be parallel planes?
4 Prove that if the circles of Intersection of two planes with a sphere are
congruent, then the planes are equidistant from the center of the
sphere.
5, Let a sphere with radius 12 be given, A segment from the center of the
sphere to a chord and perpendicular to the chord has length 8, Find the
length of the chord,
§. A sphere with center C is tangent to plane a at P. AS and CD are lines
in plane a which contain P. In what way is CP related to AJS? To CO?
Draw a figure which illustrates the given information.
7. Prove Theorem 13.13. {Hint: Give a proof similar to that of Theorem
13,12.)
8. Given a sphere S with center P as shown in the figure. HAS and CD are
chords of S which are equidistant from F, prove that
AB 5s W and LABP m Z CDP.
{Hint: Use Theorem 13.8.)
Given that AC and BD are perpendicular diameters of a sphere, prove
that ABCD is a square.
13.4 Tangent Planet 573
10. Given that AC and BD are distinct diameters of a sphere, prove that
ABCD is a rectangle.
11. Let a sphere S with center P as shown in the figure be given. C is a great
circle of S. R is a point on C and T is a point on S, but T is not on C. If
ml HPT = 60, prove that ARPT is equilateral.
12. Let a sphere H and a plane a tangent to S at point A be given. Let
plane be any plane other than a which contains A. (See the figure,)
(a) Prove that plane fi intersects sphere S in a circle C.
(b) Prove that piano /? intersects plane a m a line /.
(c) Prove that t is tangent to C at A.
(Hint: Suppose I intersects C in a second point Q, Then Q is on S
(Why?), and hence a intersects S in a second point Q. Contradiction?)
574 Circles and Spheres Chapter 13
■ Exercises 1322 refer to the sphere
S= [&$$%& + & +&mM}<
In each exercise, given the coordinates of a point, tell whether the point is
on the sphere, in the interior of the sphere, or in the exterior of the sphere.
13. (0,0,8) 18* (6, 4,2V3)
14. (0, 8,0) 19. (4,4, 8)
15. (4 3, 5) 20. (5, 0, 6)
16. (7,2,3) 21. (_2,2 X /B,0)
17. (4,5,6) 22. (4.3V$ 2)
23. Find the endpoints of three distinct diameters of the sphere
$= {{x f if,z):x* + f + J = lQa).
■ Exercises 2428 refer to the sphere with center at (0, 0. 0) and containing
the point (4 t 4* 2).
24. Find the radius of the sphere.
25. Use setbuilder notation to express the set of points an the sphere,
26. Find the coordinates of two distinct points that are on the sphere.
27. Find the coordinates of two distinct points that are in the exterior of
the sphere.
2& Fiod the coordinates of two distinct points that are in the interior of
the sphere.
29. challenge PROBLEM. Theorem 13.14 could be called a restatement
of part 2 of Theorem 13.10* Complete the proof of the following restate
ment of part 3 of Theorem 13.10.
The intersection of a plane and a sphere is a circle whose center
is the foot of the perpendicular from the center of the sphere to the
plane if and only if the foot of the perpendicular is in the interior of the
sphere.
30.
13.4 Tangent Plane* 575
Proof; 1 ,et a sphere S with center O and radius r and a plane a be
given. Let P he the foot of the perpendicular from O to a as shown in
the figure. There are two things to be proved,
(a) If OP < r, then or H S is a circle G with center P.
(1>) If a n S is a circle with center F, then OF < r.
Let X be any point of the intersection of a and S. To complete the proof
of (a) you need to show that FX is a constant for all points X in the in
tersection of a and 5. To complete the proof of (b) you need lo show thai
if FX is a constant for all points X in the intersection of ct and 8, then
OP<r.
CHALLENCF problem. Complete the proof of the following restate
ment of part 1 of Theorem 13.10.
The intersection of a plane and a sphere is the empty set if and only
if the foot of the perpendicular from the center of the sphere to die
plane is in the exterior of the sphere
Proof: Lei a sphere S with center O and radius r and a plane « be
given. Let P be the foot of the perpendicular from O to <* as shown in
the figure
There are two things to be proved.
(a) If OP > r t then all points of « are in the exterior of the sphere.
(b) If all points of « are in the exterior of the sphere, then OP > r,
31. challenge problem. See the proof of Theorem 13,2. Prove Theorem
13.10 in a similar way using an xipcoordinate system.
576 Circle and Spheres
Chapter 13
13,5 CIRCULAR ARCS. ARC MEASURE
Thus far in this chapter we have treated circles and spheres in a
similar maimer. In the remainder of this chapter, we limit ourselves to
the consideration of topics relating to circles only. The reason for this
is that the treatment of the corresponding topics for spheres is too com
plicated to consider in a first course in geometry. We begin with some
definitions.
Definition 13.12 An angle which Is coplanar with a circle
and has its vertex at the center of the circle is called a central
angle.
Figure 13S6
In Figure 1326, P is the center of Ihe given circle and AT) is a di
ameter. LAFB is a central angle. Name three more central angles
shown in trie figure.
Definition 13.13 (See Figure 1327.) If A and B are distinct
points on a circle with center P and if A and B are not the end
points of a diameter of the circle, then the union of A, B, and
all points of the circle in the interior of LAPB is called a
minor arc of the circle. The union of A, B, and all points of the
circle in the exterior of LAPB is called a major arc oi the
circle. If A and B are the endpoiuts of a diameter of the circle,
then the union of A, B, and all points of the circle in one of the
two halfplanes, with edge AB t lying in the plane of the circle
is called a semicircle.
13,5 Circular Arcs, Arc Measure 577
^i ic."n
Figure 1027
From Definition 13.13, an arc of a circle is either a minor arc, a
major arc, or a semicircle. The points A and B in Definition 13.13 are
called the endpoints of the arc.
Notation. We may denote an arc with endpoints A and B by the sym
bol AjB, which is read "arc AB." However, it should be noted that the
symbol AB is ambiguous unless the word "minor" or the word "major"
is used in connection with the symbol. For example, in Figure 1327,
we may speak of the minor AjB or the major Aft* Also, semicircle AB
is ambiguous since there are two semicircles with endpoints A and B.
One way to avoid confusion as to which arc is meant is by choosing an
interior point of the arc in question (tliat is, a point of the arc other
than its endpoints) and using this third point in naming the arc. Thus,
in Figure 1328, w© may speak of minor AXB or simply AXB. Similarly,
we may speak of major A YB or simply A YB, semicircle CXB or semi
circle CYB.
Figure 132S
=78 Circles and Sphtret Ciap:^ 13
It is clear that for each pair of distinct points A, B on a circle there
are two arcs which have these points as endpoints. If AB is a minor are,
we sometimes say that major AB is tho corresponding major are, or
that major AB corresponds to minor AB, or that minor AS corresponds
to major AB,
In Figure 1328 name the major arc that corresponds to minor BX.
Name the minor arc that corresponds to major ABY.
When we speak of the measure of a segment, we mean the nnm
bcr that is the length of the segment. However, this is not true of arcs.
That is, when we speak of the measure of an are, we do not mean the
"length" of the arc, since length has not been defined for anything
except segments. If AXB is a minor arc of a circle with center F, we
say that / APB is the associated central angle with respect to AXB.
The measure of a minor arc of a circle is related to the degree measure
of its associated central angle. We make the following definitions.
Definition 13, 14 II AXB is any arc of a circle with center P t
men its degree measure (denoted by mAXB) is given as
follows:
1. If AXB is a minor arc, then mAXB is the measure of the
associated central angle; that Is,
mAXB = mZ APR
2. If A XB is a semicircle then
tnAXB = ISO.
3. If AXB is a major arc and AY& is the corresponding minor
arcs then
mAXB = 360  niAYB,
Figure 1329 shows a circle with center i J . If
mlAPB = 50,
then
and
mAXB = 50
mXyS = 360  50  310.
13.5 Circular Arci, Arc Measure
57*3
Figure 1319
If BC is a diameter of the circle, what is mCXB in the figure? What is
mCYB?
Hereafter we shall call mAXB simply the measure of arc AXB with
the understanding that wc mean the degree measure of the arc. Is the
measure of a minor arc always less than 1 80? Why? Ts the measure of
a major arc always greater than 180? Why? Can the measure of an arc
be zero? Why?
Note that the measure of an arc does not depend on the size of the
circle which contains the arc. Figure 1330 shows 'Concentric circles
with center P. with
mCYD = mAXB = mLAVB = mlCPD = 35.
Figure 1330
Given three points A, B, C such that B is between A and C, we
know that
AB 4 BC = AC.
Suppose we are given an arc ABC (that is, B is a point, but not an end
point, of AC). It seems reasonable that
inAB + JtiBC = rnAFC.
We state this as our next theorem.
5 BO Circtas and Spheres Chapter 13
THEOREM 13.15 {Arc Meamre Addition Theorem) If A, B, C
are distinct points on a circle, then
mXlid = mAB + mBC
Proof; Followi ng is a plan for a p roof considering seven possi ble cases
as shown in Figure 133 1 . In each case, V is the center of the circle and
the assertion of the theorem follows from the listed equations. This
plans a complete proof since there are no other cases.
c A\
U.C J
Ciae2
Cosu U
Cut?
Cast I Quti
Hgur* 1331
Case 1, ABC is a minor arc. Then AB and BC are minor arcs,
mABC = mlAVC
mAB = ml AVB
mlRm ml BVC
m I AVC = m I AVB + ml BVC.
Case 2. ABC is a semicircle. Then A B and BC are minor arcs,
mABC = 180
mAB = mlAVB
mBC ml BVC
180 = m I AVB + m L BVC.
3. ABC is a major arc, and AB and BC are minor arcs. Then
mABC = 360  mlAVC
mAB = m I AVB
mBC = ml BVC
360 = ml AVB + ml BVC + mlCVA.
13.5 Circular Arcs, Arc Manure a 81
Case 4. ABC is a major arc, AB is a minor arc, and BC is a major arc.
Then
mABC = 360 mZ.AVC
mAB = mLAVB
mBC = 260 mLBVC
mLBVC = mLBVA + mLAVC.
Goa& 5. AJ3C is a major arc, AB is a minor arc, and BC is a semicircle.
Then
mABC = 360  mLAVC
mAB = mLAVB
m£C= 180
180 = mZAVB + mLAVC
Case 6. ABC is a major arc, AB is a semicircle, and BC is a minor arc.
Then
mABC = 360 mLAVC
mAB = 180
mBC = mLBVC
180 = mLAVC + mLBVC.
Case 7. ABC is a major arc. AB is a major arc, and BC is a minor arc.
Then
mABC = 360 mLAVC
mAB = 360 mLAVB
mBC = mLBVC
mLAVB s mLAVC + mLCVB.
Note in the situation of Theorem 13, 15 that if D is a point of the
circle not on ABC, then
mCD/1 = 360  mABC
and
mAB + mBC + mGDA = mA5c + mCDA = 360.
In other words, if A. B, C are three distinct points on a circle that par
titions the circle into arcs, AB, BC, CA, intersecting only at their end
points, then
mAB + mBC + inCA =
582 Circles and Spheres Chapter 13
This idea may be extended to any numl>er of points on a circle, as in
the following corollary.
COROLLAR Y I.'i, 15,1 U A i, A 2t . , . , A E are n distinct points on
a circle that partition the circle into n arcs
AjA 2 + AvAz* • . . * A„_iA B> A^Ai
that intersect only at their endpoints, then
tnAiA 2 + mAvAs + ■ • * + mA n _iA n + ntA Ji A 1 = 360.
Proof: Following is Lhe proof for n = 5, The proofs for other values
of n arc similar. By repeated application of Theorem 13.15, we get the
following:
mA L A 2 + mAzA s 4 mAfiA* 4 mA^Aa + mA^A L =
mAiA'jAy f mAaA4 + 771A4A5 + mA$A l =
mAiA^A* 4 rnA+As  mA$A\ =
mAiA 2 A 5 r mAgAi = 360.
EXERCISES 13,5
■ Exercises 17 are on the proof of Theorem 13,15. In each exercise, show
how to derive the assertion oJ the theorem from the listed equations for the
given case.
1. Case 1 5. Case 5
2. Case 2 6. Case 6
3. Case 3 7. Case 7
4. Case 4
8. In the figure, mABC = 240 and mBA'C = 100. Find mAZB and ntAYC.
A ^ ^ Z
13.5 Circular Arcs, Arc Measure 583
Exercises 920 refer to Figure 1332. In the figure, P is the center of the
circle; A, B, C, are points on the circle; APB, mBC — 50; and mBD =s
110.
no
Figure I&32
9, Name five minor arcs determined by points labeled in the figure.
10. Kind tile measure of each of the minor arcs named in Exercise 9.
11. Name the five major arcs corresponding to the five minor arcs named in
Exercise 9,
12. Find the measure of each of the major ares named in Exercise 11.
13. Name two semicircles.
14. Which theorem justifies the conclusion thai mCBD = 160?
15. Find m LBFC.
16. Find m L LTD.
17. Copy Figure 1332 and draw A£, BC t and FC. Find mlBAC, How
does m L BAC compare with mBC?
ia Show that mlACB = 90.
19. Draw A~D, BD, and PD on your copy of Figure 1332. Show that
ml BAD = [mBlX
20. Find m£ADB.
Exercises 2123 refer to Figure 1333. In the figure, Pis the center of the
circle; A, B, C are points on the circle; and APB.
21. Copy Figure L333 and draw FC.
Prove thiil ml BAC = \mBC,
22. Prove thai in L ABC = fynAC.
23. Prove that A ABC is a right
triangle.
Figui* 1*33
584 Circles and Spheres
24. CHAIJ.ENGE PROBLEM. Ill the KgUTe,
Pis the center of the circle; A t B,
C, D arc points on the circle;
and AFB. Prove that
m£CAD = %mCBD.
13.6 INTERCEPTED ARCS, INSCRIBED ANGLES, ANGLE MEASURE
In the discussions that follow we shall he concerned with angles in
scribed in an arc of a circle and about arcs of a circle that are inter
cepted by certain angles. In Figure 1334a, £ABC is inscribed in ABC
and t ABC intercepts AXC In Figure l&34b, £PQR is inscribed in
PQR and Z FQR intercepts FYR.
Figure 1334
None of the angles shown in Figure 1335 is an inscribed angle, but
each angle intercepts one or more arcs of a circle. In Figure 1335a,
AAFB intercepts AXB. In Figure 1335h, Z R ST intercepts RYS, In
Figure 13.35c £AVB intercepts AKR and also CMD. In Figure
1335d, Z GEF intercepts GHF, In Figure 1335C, Z A VB intercepts
ARB and also CSB, In Figure 1335f t £AVB intercepts Ai5i and also
AGB.
13.6 Intercepted Arcs, Inscribed Angles, Angle Measure 585
Figure 1335
On the other hand. L LKP
shown in Figure 1336 does not
intercept an are of the circle.
Figure 1UH0
586 Circlet and Spheres
Chapter 13
Our formal definitions of an Inscribed angle and of an intercepted
arc follow.
Definition 13.15 An angle is said to l>e inscribed in an arc
of a circle and is called an inscribed angle if and only if both
of the following conditions are satisfied ;
1 , Each side of the angle contains an endpoint of the arc.
2. The vertex of the angle is a point, but not an endpoint, of
the arc.
Explain why each of the angles shown in Figure 1335 fails to be an
inscribed angle. Explain why each of the angles shown in Figure 1334
is an inscribed angle. Draw a picture of an angle inscribed in a
semicircle.
Definition 1X16 An angle is said to intercept an arc of a
circle and the arc is called an intercepted arc of the angle if
and only if all three of the following conditions are satisfied:
1. Hie end points of the arc lie on the angle.
2. Each side of the angle contains at least one endpoint of the
arc.
3. Each point of the arc, except its endpoints, lies in the in
terior of the angle.
You should check to see that each of the figures shown in Figure
1335 satisfies all three of the conditions stated in Definition 13.16.
Which of the tlircr conditions stated in Definition 13.16 are not satis
fied by the figure shown in Figure 1336?
Figure 1337 shows two angles inscribed in the same arc. (Of
course, the angles intercept the same
arc) Name the arc in which both
angles, LABD and £ ACD, are in
scribed. Name the arc that both of
these angles intercept, it appears
that Z ABD and LACD in the figure
are congruent. (Measure each of
them with your protractor,) That
they actually are congruent is a cor
ollary of our next theorem.
Figure 1337
13.S Intercepted Arcs, Inscribed Angles, Angle Measure 587
THEOREM 13.16 The measure of an inscribed angle is onehalf
the measure of its intercepted are.
Proof: Let a circle with center V be given and let Z ABC be inscribed
in ABC. Then the intercepted arc is AC* We must prove mLABC =
JmAC There are three possible cases as suggested in Figure 1338.
Casel. VisonZA^C.
Case 2. V is in the interior of Z ABC,
Case 3. V is in the exterior of /ABC.
Figure 1338
Proof of Case I: Since V is on Z ABC, then either AB or BC is a di
ameter of the given circle. Suppose BC is a diameter as in Figure
I33&L Draw VA Then AAVB is isosceles (Why?) and mLA =
m it ABC /.AVC is an exterior angle of AAVB, so by Theorem 730,
m/AVC = mlA + mLABC
Since m£A = m£ABC r
nUAVC = mLABC + mZA3C,
or
But
Therefore
2?n£ABC = mlAVC,
m£AVC = mAC (Why?)
2mZABC= mAC
mLABC = ^niAC.
This completes the proof of Case 1 .
Proof of Case 2: V is in the interior of /.ABC as shown in Figure
1338b, so B Vis between BA and BC. Let B' be the point where opp VB
intersects the circle; then BW is a diameter and B' is an interior point
5SB Circles irwj Spheres Chapter 13
of the intercepted arc AC. It follows from Case I that
m I ABB' = $mAB' and ml B'BC = ^mBC
Adding, we get
ml ABB' + m/B'BC = \mAB> + ^nB~C,
or
ml ABB' + mlWBC = J(w*dB' + mB'C).
But
ml ABB' + ml B'BC = ml ABC (Why?)
and
mAB' + miPc = wtAC : Why?)
Therefore
mlABC= fytiAC
and the proof of Case 2 is complete. The proof of Case 3 is assigned as
an exercise.
Recall how congruent segments and congruent angles are defined
in terms of their measures. How would you define congruent arcs?
Write what you think is a good definition of congruent arcs. Turn back
to Figure 13*30. Does your definition allow the conclusion that arcs
AXB and CYD of Figure 1330 arc congruent? If it does, you should
reword your definition so that it excludes this conclusion. Compare
your definition with the following.
Definition 2,127 Two arcs (not necessarily distinct) are
congruent if and ony if they have the same measure and are
arcs of congruent circles.
Notation H arcs AB and CD are congruent, we write AB = CD.
Note mat we cannot say that two arcs arc congruent unless we
know that they have the same measure and we know that they are in
the same circle or that they are in congruent circles.
The following three corollaries are important consequences of The
orem 13.16, Their proofs are easy and are assigned as exercises.
COROLLARY l&W.l Angles inscribed in the same are are
congruent.
COROLLARY 13.16.2 An angle inscrilied in a semicircle is a
right angle.
13.6 Intercepted Are*, Inscribed Angles, Angle Measure 539
COROLLARY 13.16.3 Congruent angles inscribed in the same
circle or in congruent circles intercept congruent arcs.
Figure 1339 shows two distinct purullel lines intersecting a circle.
We call an arc whose endpoints arc on the lines, one endpoint on each
of the lines and all of whose interior points are between the lines, an
intercepted arc. Thus, in Figure 1339, lines An and CD intercept arcs
AXC and BYD. Are these two arcs congruent? They are congruent by
our next theorem.
Figure 1339
THEOREM 13.17 If two distinct parallel lines in the plane of a
circle intersect that circle, they intercept congruent arcs.
Proof: Let distinct parallel lines / and in intersect a given circle in
arcs AXC and BYD. There are three possible cases as suggested in
Figure 1340.
B B
Figure 1340
Case L Both I and m are secant lines as shown in Figure 1340a.
Case 2. The line if is a tangent line and the other line m is a secant
line as shown in Figure 13~40b.
Case 3. Both I and m are tangent lines as shown in Figure 134Qc,
590 Circles arui Spheres Chapter 13
Proof of Case 1: Since / and m are secant lines, I intersects the circle
in distinct points A and B, and m intersects the circle in distinct points
C and D. (See Figure 1340a.) We must prove that AXC m BYD. Draw
B€. Then
and
But
Therefore
or
mlBCD =
ml ABC = $mAXC.
mlABC = m L BCD. Why?
\mAXC s JmBY/3
mAXC = mBYD.
AXC B BYD. Why?
This completes the proof of Case 1 of Theorem 13,17, The proofs
of Cases 2 and 3 are assigned as exercises,
EXERCISES 13.0
Exercises 117 refer to Figure 134L In the figure* P is the center of the
circle: A, B, C, D arc points on the circle; AC is a diameter; mAD = 100;
and mBC b 40.
Kipiw 1341
1. Kamc four inscribed angles and name the arc each intercepts.
2. Name the five minor arcs determined by points labeled in the figure.
3. Name the five major arcs corresponding to the five minor arcs named
in Exercise 2.
4. Name L\vr> semicircles.
5. Find the measure of each of the minor arcs named in Exercise 2,
CJ. Find the measure of each of the major ares named in Exercise 3.
13.6 Intercepted Arc*, Inicribed Anglei, Angle Measure
7. Find the measure of each of the angles ZA, Z B, L C t and Z0.
8. Find the measure of £CFD,
9. Find the measure of LABC.
10. Name an angle thai [S congruent to ZB,
11. Name an angle that is congruent to LD,
12. Prove that AD 1 CD.
13. Find the measure of Z DCB.
14. II AC intersects BD at K, name a pair of similar triangles,
15. If AC intersects BD at E, prove that m / DEC = %{mCD + mAU).
16. Prove that AADE  AW(7
17. Prove that DE ■ £# = AE  EC,
591
18. With each chord of a circle which is not a diameter there are two asso
ciated arcs of the circle. One of the arcs is a minor arc and the other arc
is a major arc. The endpoints of the chord are the end points of the arcs.
Complete the proof of the following theorem.
THEOREM 13. IS In the same circle, or in congruent circles, two
chords that arc not diameters are congruent if and only if their asso
ciated minor arcs are congruent.
Proof: Using the notation of the figure, we are given two congruent
circles , C a nd C, with centers P and F, respectively. AB is a chord of
C and A'W is a chord of C. There arc two things to prove.
(a) If AB ~ A'B'. then A B tt A'B',
(b) If AB s j?B', then M a A^F.
{Hints In proving (a), show that AAPB s AAT7*' by the S.S.S. Pos
tulate. In proving (b), show that AAPJ? at AA'P'B' by the S.A.S.
Postulate.'!
592 Circles and Spheres
Chapter 13
19. Does Theorem 13.18 of Exercise 18 still hold if we replace "minor arcs"
with "major arcs" in the statement of the theorem?
20. Prove Case 3 of Theorem 13.16. (Refer to Figure 1338c.)
21. Prove Corollary 13.16.1.
22. Prove Corollary 13.16.2.
23. Prove Corollary 13.16.3.
24. Figure 1342 shows three cases of a chord AT of a circle with center P
and a tangent VTto tlie same circle intersecting at the point of taugency
V. Using the notation of the figure, the angle whose sides are rays VT
and VA is sometimes called a tangentchord angle. Complete the proof
of the following theorem.
THEORtlM 1^.19 The measure of a tangentchord angle is onehalf
the measure of its intercepted arc.
Proof: Using the notation of Figure 1 342, P is the center of the given
circle, AXV is the intercepted arc, AT'is the given chord, and VTis the
given tangent. There are three cases to consider.
Figure 1348
Case L P is on VA as shown in Figure 1342u.
Case 2. P k in the exterior of I AVT as shown in Figure 1342b.
Case 3. P is in the interior of I A VT as shown in Figure 1 342c.
Proof of Casn h (See Figure 1342a,} AV is a diameter; hence
mZ AVT = 90 (Why?) and mAXV = 180. Thus ml AVT = fynAXV.
Proof of Case 2: (See Figure 1342b.) Draw diamotcr VB.
mlBVl = imHXV, Why?
= limM +mAXV), Why?
= {niffi + tyiAX$.
Since mlBVT = mlBVA +mZ AVT, we have
ml BVA + mlAVr = tynBA + intAXV
fey the Substitution Property of Equality. Rut
13.6 Intercepted Arc*, inscribed Angles, Angle Measure 593
mlBVA = imBA. Why?
Therefore, by the Addition Property of tonality, we get
mlAVT=%mAXV,
Complete the proof of Theorem 13,19 by proving Case 3,
25. Prove Case 2 of Theorem 13.] 7.
20. Prove Case 3 of Theorem 13.17. {Hint: In Figure 1340c, prove that W
is a diameter. It will then follow that HXD and BYD are semicircles
and that BKD S BYD.)
27. A quadrilateral is said to he inscribed In u circle and is called an in
scribed quadrilateral if all of its vertices are on the circle. Prove that the
opposite angles of an inscribed quadrilateral are supplementary.
28. If the diagonals of an inscribed quadrilateral are diameters, prove that
the quadrilateral is a rectangle, (See Kxerci.se 27.)
29. Prove that the midray of a central angle of a circle bisects the arc inter
cepted by the angle.
30. The figure shows two secants intersecting in an exterior point V of a
circle. The rays VA and VC are sometimes called secantray?; and the
angle whose sides are these rays is called it secantsecant angle. Com
plete the proof of the following theorem.
THEOREM 13.20 The measure of a secantsecant angle is onehalf
the difference of the measures of the intercepted arcs.
Using the notation of the hgure t we must prove
fliZV = JrimBYD  mAXC),
We have
mZV+ ml ABC  m L BCD. why?
Therefore
mZV+ \mAXC = l 2 mBYD. Why?
Complete the proof.
594 Circles and Spheres
Chapter 13
31. An angle whose sides are a secantray and a tangentray from an exterior
point of a circle is called a secanttangent angle. Prove that the measure
of a secanttangent angle is onehalf the difference of the measures of
the intercepted arcs,
(Hint: Using the notation of the figure you must prove that
mZ V = fynBYC  mAXC).
Use Theorem 13.19 stated in Kxercisc 24. Also see Exercise 3CLJ
32. An angle whose sides are two tangentrays from an exterior point of a
circle is called a tangenttangent angle. Prove that the measure of a
tangenttangent angle is onehalf the difference of the measures of the
intercepted arcs.
(Hint: Using the notation of the figure you must prove, that
m£V= limAYB  mAXB)
Use Theorem 13.19 stated in Exercise 24. Also sec Exercise 30.)
33. Copy and complete the following theorem which combines the state
ments of Theorem 13.20 (see Exercise 30) and Exercises 31 and 32 into
a single statement.
13.6 Intercepted Arc*, Inscribed Angles, Angle Miaiure 595
THEOREM 13.21 Jf an angle has its vertex in the exterior of a circle
and if its sides consist of two secantrays, or a secantray and a tungeut
ray, or two tangentrays to the circle, then Hie measure of the angle
34. Complete the proof of the following theorem.
THEOREM 13.22 The measure of an angle whose vertex is in the
interior of a circle and whose sides are contained in two secants is one
half the stun of the measures of the intercepted arcs.
(An angle such as the one described in Theorem 13.22 is called a
chordchord angle.
Using the notation of the figure, it is given that S3 and CD are
secants intersecting at E, a point in the interior of the circle. Prove that
ml DEB = ^rnDYB + mAXC).
Exercises 3544 refer to Figure 1313. In the figure, VA and VC are secant
rays of the circle with center P, VE is a tangentray, chords A~D and BC
intersect at F, and BPC. If the measures of arcs AB, BD, £*£, and AC are
as shown, use the results of Exercises 30, 31, 32, and 34 and other theorems
proved in this section to find the indicated measure in each exercise.
35* m£AVG
36. mCE
37. mlCVE
38. mLAVE
39. mlBFD
40. mlAFB
4L mLDEG
42. ml BEG
43. ml CAB
44. mlBEA
Figure 1*43
596 Circlet and Spheres Chapter 13
45, challenge PROBLEM* In the figure, circles C and C with centers P
and F t respectively, are tangent internally at R and Circle C contains P.
If RH is any chord (with one endpoint atfi) of C and if H? intersects C
at Af, prove that M is the midpoint of KS.
13.7 SEGMENTS OF CHORDS, TANGENTS, AND SECANTS
If two distinct chords of a circle intersect at an interior point of the
circle, the point of intersection together with the midpoints of the
chords determine four distinct segments (other than the chords) which
are subsets of the chords. For example, in Figure 1344, chords AB and
CD intersect at P. The four segments to which we refer are segments
Ai\ PB, t J", and PD. I 1 is eas> to prove that the protlut 1 \>i 1 he lengths
of the two segments on AB is equal to the product of the lengths of the
two segments on CD.
^ +^ Figure JIM4
THEOREM 13,23 If two chords of a circle intersect, the product
of the lengths of the segments of one chord is equal to the product
of the lengths of the segments of the other.
Proof: Using the notation of Figure 1344, we are given a circle with
chords AB and CD intersecting at P. We are to prove that
APPB = CPPD,
137 Segment* of Chords, Tangents, and Secants
We draw AD and EC. Then
597
LA=z LC
£Dm LB.
(Why?)
(Why?)
and
Therefore
AADP  ACBP
by the A, A, Similarity Theorem. It follows that
(AP, PD) = (CF, PB)
and that
APPB = CP*PD.
This completes the proof of Theorem 13,23.
Figure 1345 shows a line VT tangent to a circle ut Tand a secant
FA intersecting the circle in points A and B. in die statements of our
next three theorems, we refer to segments such as VT, PB, and FA.
Figure 1345
Therefore it is convenient to have a name for each of them. We make
the following definition.
Definition 13.18 If Vand T are distinct points and if the
line VT is tangent to a circle at T, then the segment VT is
called a tangentsegment from V to the circle. If secant PA
intersects a circle in points A and B such that A is between P
and B t then the segment FB is called a secantsegment from
P to the circle and the segment 1*A is called an external
secantsegment from P to the circle.
The proof of our next theorem is Exercise 37 of Exercises 13.3.
598 Circles and Spheres
Chapter 13
THEOREM 13.24 The two distinct tangentsegments to a circle
with center O from an external point P are congruent and the angle
whose vertex is P and whose sides contain the I wo tangent
segments is bisected by the ray PO.
Proof: Using the notation of Figure 1346, given a circle with center
O, tangentsegments PA and PB, and ray PO, we are to prove that
J5i fiS FE and that PO bisects LAPB. Give reasons for the statements
in the proof when asked,
Z CMP and Z OBP are right
angles. YVhy? Therefore A GAP
and AOBP are right triangles,
OA^OB :Why?)
and
FO^FO,
Therefore
A OAP ^ A OBP. { Why?)
It follows that PA Si PB and that ZAPO ^ ZBPO. To complete the
proof we need to show that O is in the interior of LAPB.
The union of the given circle and its interior is a convex set. Call
it S. If Q is any point of S, then OQ < OA, If R is any point of PA, ex
4*
ccpt A, then OR > OA. Therefore S does not intersect PA except at
A. Therefore all of S except A lies on one side of PA. In particular, O
lies on the Bside of PA. Similarly, it may be shown that () lu.s mi the
A 'Side of PB. Therefore O is in the interior of Z.APB and since
Z.APO « ZjBPO, it follows that PO bisects ZAPB.
17JEOBEM f.3.25 The product of the length of a secai itsegment
from a given exterior point of a circle and the length of its external
secantsegment is the same for any secant to the given circle from
the given exterior point.
Figure 1346
Figure 1347
13.7 Segments of Chords, Tangents, and Secants 599
Using the notation of Figure 1347, given secantsegments FE and
FD and external secantsegments J5t and JPG, we are to prove that
PR • PA = TO ■ PC. Plan: Prove APCB ~ APA D.
Proof: Assigned as an exercise.
Our next theorem gives us still another relation between the prod
ucts of the lengths of certain segments. Figure 1348 shows an exterior
point P of a given circle and a tangent to the circle at C which contains
the exterior point P. If t is any secant to the given circle which contains
P and intersects the circle in points A and B, we can prove that the
product PA * PB is the same as the product PC ■ PC, or (PC) 2 .
Figure 13*S
THEOREM 13.26 Given a tangentsegment PC from P to a circle
at C and a secant through P intersecting the given circle in points A
and B, then
PA • PB = (PC) 2 ,
Proof: Using the notation of Figure 1348, we are given a tangent
segment TC from an exterior point P to the circle at C PB is any secant
segment from P and intersecting the given circle in points A and B. We
are to prove that PA*PB = (PCp.
L mlPCA = ^mAC
2. m/,B = Jmulfi
3. mLPCA = mLB
4. ZPCA = LB
5. ZPs ZP
6. APCA ~ APBC
7. (PC, PA) = (PB, PC)
8. PA*PB = (PQ 2
Benson
1. The measure of a tangent
chord angle is onehalf tile
measure of the intercepted arc.
2. Why?
3. Why?
4. Why?
5. Why?
6. Why?
7. Why?
8. Why?
600 Circles and Spheres Chapter 13
Tt follows from Theorem 13.23 that if P is any point in the interior
of a circle, the product PA • PB remains unchanged for any chord 7lB
which contains P. Also, if P is a point in the exterior of a circle as in
Theorems 13.25 and 1 3.26, the product PA  PB remains unchanged for
any secantsegment from V and intersecting the circle in points A and
B, or for any tangentsegment from P to the circle. Theorems 13,25
and 26 suggest that the value of the "unchanged product" is the square
of the length of the tangentsegment from P to the circle. This seems to
have no significance for Theorem 13.23. Figure 1349 suggests (with
a bit of help from Pythagoras) that (OP) 2 — r 2 exists even when P is
inside the circle. As we shall see, the idea of {OP} 2 — r 2 as the value of
die "unchanged product*' is what relates Theorem 13,23 to Theorems
13.25 and 13.26.
Figure 1349
Definition 13,19 Given a circle S with center O and radius
r y and a point P in the same plane as 5, the power of /' with
respect to S is (OF) 2 — r 2 .
Given a circle S with center and radius r, a point P coplanar with
S is in the interior of S, on S, or in the exterior of S, according to
whether OP < r, OP = r, or OP > i\ hence according to whether
(OP)'* < r'^iOF} 2 = i* or {OP) z > r 2 , and hence according to whether
the power of P with respect to S is negative, zero, or positive. If an
iycoordinate system is set up in the plane of S with its center O as
origin, then the power of P with respect to S is
(OF)*  r* = x 2 + if  r 2
and
interior of S = {(*, y) : x 2 + if  r* < 0}
S = {(*, y) : x 2 + y*  r 2 = 0}
exterior of S = {(%, y) i x 2 + y 2 — r 2 > 0).
Compare this with the rep rose utations using setbuilder notation in
Figure 139.
13.7 Segments of Chords, Tangents, ami Secants 601
We are now ready for the theorem that relates Theorem 13.23 to
Theorems 13,25 and 13.26.
THEOREM 13.27 Let a circle S with center O and radius r be
given. Let P be a point in the plane of 4? and let p be the power of P
with respect to S. If a line through P intersects S in points A and
B t then
1. PAPB= p if Pis inside S, and
2. PA • PB = p if P is on S or outside S.
Proof; Our proof is by cases. In Cases 1 and 3, EF is the diameter of
S such that EF contains P,
Case I.
PA • PR = FE • PF (Why?)
= (r  Of)(r + OP)
= r 2  (OP) 2
= P
Casel
J» = A
Case 2a.
PA • PB = ■ PB
=
= (Of) 2  r 2
Case 2fr.
PA ■ PB = ■ •
=
= (Oi^ 2  r*
Case 2a
Case 2b
Cow 2c.
PAPB = PA •
■
= (OP)2  r2
= p.
^■^ /i
Case 2c
602 Circles and Spheres
Chapttr 13
Cases 3a and 3b.
FAPB = PEPE (Why?)
= {OP + r)(OP  r)
= (Of) 2  r*
O.iao 3b
The theorems wc have proved in this section enable us to do many
numerical problems.
Example 1 In Figure 1350. CD = 38, CE = 2Q S and BE ^ 24. Find
DE and AE.
Solution: DE = CD  CE
= 38  20 m 18.
Let AE = %; then, by Theorem 13.23,
*•££ = CEDE.
Thus
*24 = 2018
or
24* S 360 Figure 1340
and
x= 15.
What is the power of £ in this example? Is the power of E a positive
number or a negative number?
Example 2 In Figure 1351, PB = 70 and PC = 40. Find PA.
Figure UMQ
13,7 Sagment* of Chords, Tangents, and Secants 603
Solution: By Theorem 13.26, (PA) 2 = PB • PC. Therefore
(PA) 2 = 7040.
Thus
and
(PA) 2 = 2800,
(FAV = 400 ■ 7.
FA = V4007,
PA = 20 yT
What is the power of P in Example 2? Is the power of P a positive
number or a negative number?
Exampte 3 In Figure 1352, FB = 78, AB a 26, and PD = 82.
Find CD.
D
Figure 1352
Solution: First we need to find PA and PC. Why? We have
PA = FB  AS = 78  26 = 52,
By Theorem 13.25,
PDPC = PR PA.
Therefore
82PC=7852
PC = 7852 m 49 19
82 41
Thus
CD = FD  PC a 82  49$ = 32f
What is the power of F in Example 3?
604 Circles and Spheres
Chapter 13
EXEBCISES 13.7
I, Prove Theorem 13.25. (liefer to Figure 1347.)
Exercises 25 refer to Figure 1 353 which shows two intersecting chords of
a circle.
Figure 1353
2L UCE= 7 and DE = 9, find the power of E.
3. If AB = 24, BE = 8, and C£ = 12, find DE and the power of E.
4. If CE = r, DE m 8. AE = 12, and BE = x  2 f find CE f BE, and the
power of E.
5. If AE = x BE = x  4, CE = 4, and ED = 16  x, find AE, BE, £D,
and the power of E. 1 low are A"U and CD related?
Exercises 614 refer to Figure 1354 which shows two secantraj's and a
tangentray from an exterior point of a circle. If an answer is an irrational
number, put it in simplest radical form. (For example, \/32 = 4 \/2 i» sim
plest radical form. )
Figure 1334
6. If PE =16,PD= 10, and PB = B t find PC.
7. Find the power of P in Exercise 6,
8. Find PA in Kxercise 6,
9. If ED = 9, DP = 12, and PC m IS, find BC.
13.7 StgmanU of Chords, Tangents, and Secants 605
10, Kind the power of F in Exercise 9,
11, Find PA in Exercise 9,
12, If PA = 16, FB = 10, and PE = 24, find DE, PC, BC, and the power
ofP<
13, If PA m x, PE = 50, PD = 32, and PC = x + 20, find PA, PC, PB,
and BC,
14, Find the power of P in Exercise 13,
15, The figure shows two circles intersect
ing at Q and H, a tangentray from V
to the larger circle at P, a tangentray
from V to the smaller circle at S. and
a lecantray from V intersecting the
two circles in points Q and B. Prove
that VP = VS. (The line QR in the
Ggyre is called the radical axis of the
two circles. It is the set of all points of
like power with respect to both circles.)
16, The figure shows two circles tangent
externally at T and vT is a common
tangent, A secantray from V inter
sects the larger circle in points A and
B, and a secantray from V intersects
the smaller circle in points R and S,
Prove VA  VB = VR  VS.
17. Given the figure in Exercise Hi Prove that AVAR ~ AVSB.
18. Given that the sides of quadrilateral ABCD are tangent to a circle at
points H, E, /, P, as shown in the ligurc, prove that
AB + CD = AD + BC
B
606 Circles and Spheres
Chapter 13
19. If a common tangent of two circles docs not intersect the segment join
ing their centers, it is called a common external tangent If it does in
tersect the segment joining, their centers, it is called a common internal
tangent. In the figure, Kb is a common external tangent of the two
circles with centers P and Q> PR = 21, QS = 6, and PQ = 25. Find AS.
(Hint: Draw $T 1 PK at Z)
20. Cive reasons for steps 2 to 8 in the proof of Theorem 13.26.
21. Complete the proof of the following theorem.
THEOREM If A and B are distinct points on a circle and if it is any
point between A and B, then ti is in the interior of the circle.
Proof: Set up an xyeooiduiate system with the origin at the center of
the given circle and with the XftXis the perpendicular bisector of AH
as shown in the figure. Let r be the radius of the circle. Then there is a
number a such that r<a<r and
A& = {(x, y) : x = a).
Then the endpoints of AB arc (a, \fF~~ a 2 } and (a, — yr 2  a 2 }, and
a point R{ a t tj) is between A and B
if and only if  \/^  a 2 < y < V^  <&,
if and only if y 2 < r 2 — a 2 .
Complete the proof hy showing that OR < r and hence that it is in
the interior of the circle.
Chapter Summary 607
22. Let line I be a tangent to a circle ul T, Prove that all points of the circle,
except 7', are on one side of I in the plane of the circle. {Hint: Let A
and B be anv two distinct points of the circle such that A *£'£ mid
B ^= 1\ and suppose that A and B are on opposite sides of I. Then there
is a point of I between A and B. Why? Therefore I intersects the in
terior of the circle. Why? See Exercise 21. Contradiction?)
23, challenge FROBliKM. Given a right triangle, APQR, with the right
angle at H, let C bo the circle with center at P and radius PR as shown
in the figure, Then £Jft is a tangentsegment (Why?) and the circle in
tersects QP in two points S and T. Why? Use Theorem 1 3.2fl and prove
that
{QFf = (QRf + (RF)Z.
Thus you will have given another proof of the Pythagorean Theorem.
CHAPTER SUMMARY
The following terms and phrases were defined in tills chapter. Be sure
that yon know the meaning of each of them.
CIRCLE
SPHERE
RADIUS OF CIRCLE (SPHERE)
DIAMETER OF CIRCLE
(SPHERE)
CONCENTRIC CIRCLES
( SPHERES)
CHORD OF CIRCLE (SPHERE)
TANGENT LINE TO CIRCLE
TANGENT CIRCLES
TANGENT PLANE TO SPHERE
SECANT
CONGRUENT CIRCLES
(SPHERES)
INTERIOR OF CIRCLE
(SPHERE)
EXTERIOR OF CIRCLE
(SPHERE)
GREAT CIRCLE
CENTRAL ANCLE OF C1RCI £.
MINOR ARC OF CIRCLE
MAJOR ARC OF CIRCLE
SEMICIRCLE
DEGREE MEASURE OF ARCS
INSCRIBED ANGLE
INTERCEPTED ARCS
CONGRUENT ARCS
TA NGENT— SEGM ENT
SECANT— SEGMENT
EXTERNAL SECANT—
I GMENT
POWER OF A POINT
608 Circles and Spheres Chftptor 13
There were 27 numbered theorems in this chapter. You should road
them again and study them so that you understand what they say. The fol
lowing are a list of a few of the more important theorems concerning circles.
You should know the corresponding theorems with regard to spheres where
applicable.
THEOREMS 13.3 and 13.4 I ^sl a circle and a line in the same plane
be given. The line is tangent to the cirde if and only if it is perpen
dicular to a radius at the outer end of the radius.
THEOREM 13.5 A diameter of a circle bisects a chord of the circle
other than a diameter if and only if it is perpendicular to the chord.
THEOREM 13,8 Chords of congruent circles are congruent if and
only if they are equidistant from the centers of the circles,
THEOREM Hi. 1ft The measure of an inscribed angle is onehalf the
measure of its intercepted arc.
REVIEW EXERCISES
In Exercises 122, refer to the circle C* with center P shown in Figure 1355,
Assume that all points in the figure are where they appear to be ia the plane
of the circle. Copy the statements, replacing the question marks with words
or symbols that best name or describe the indicated parts.
Firuit 1355
1. AB is called a \Tj of *h e circle.
2. PE is called a T of the drcle.
3. AT) is called a [TJ of the circle, KB could also be called a (JJcl the circle.
4. AB is called a [T].
5.IfSnc={T) l then fi? is called a [7] to the circle at '/'.
Review Exercises
609
6. If TK n C = {T}, then TK is to FT at I,
7, Those points named in the figure and which arc on the circle are jTJ
S. Those points named in the figure and which are points in the interior
of the circle are [?}
9. [?] is a point in the exterior of the circle.
10. L ADF is an [7] angle.
11. L BAD is [?] in arc BAD.
12. Z ADF intercepts arc Tj.
13. DTE is a \J\ arc of the circle.
14. DAF is a J} arc of the circle.
15. AM} is a [?J.
W. mZBAI? is mB7).
17. L EFT is a [T] angle.
18. m/T=[7J.
19. The power of <? is [TJ.
20. Hie power of K is g$
22. Tlic power of A is \T\ r
22. A aJ0 Is similar to AQJ
In Exercises 2329 refer to the sphere S with center Q shown in Figure
1356, Copy the statements, replacing the question marks with words or
symbols that best name or describe the indicated parts.
Figure 1356
23. QR is a [3 of the sphere.
24. If V t Q, Tare collincar, then Vf is a [?] of the sphere.
25. If W, Q, R are coQmear, then the circle with center Q and containing
points W t F, R is a of the sphere,
26. RV is a [Tj of the sphere.
27. HP is a [JO
28. If a n S = {T), then a is Q] to S at T.
29. If a H S = {T}, then Qf is [7] to a at T,
9ft Find the radius of a circle if one of its chords 16 in, long is 6 in. from
the center of the circle.
610 Circles and Spheret Chapter 13
31. How far from the center of a circle with radius 16 is a chord whose
length is 24?
32. A circle C and a line Hn tin ayplane are given by
C = {{ar, y) : a* + y2 = 36}
(a) Write the coordinates of six points that are on the circle.
(b) Write the coordinates of three points that are on the line.
(c) Write the coordinates of two points that are on the circle and on the
line,
33. A sphere S in an xyscoordinate system is given by
S = {{*> y, «) : *• + g* + # m 81}.
(a) Write the coordinates of eight points that are on the sphere.
(b) Write the cooreli nates of two points that are in the interior of the
sphere.
(c) Write the coordinates of two points that are in the exterior of the
sphere.
(d) Write the coordinates of two points on the sphere that are endpoints
of a diameter of the sphere and are not on any of the three coordi
nate axes.
34. Write an equation of a sphere with center at the origin of an xtpcoor
dinate system and which contains the point (4, —2, 5). What is the
radius of this sphere?
35. Let a sphere with radius 10 be given. A segment from the center of the
sphere to a chord and perpendicular to the chord has length 4, Find the
length of the chord.
In Exercises 3o 45 refer to die circle with center P shown in Figure 1357.
Given the notation of the figure and the degree measures labeled in the
figure, find the measure asked for in each exercise. You may need to refer
to the theorems stated in Exercises 24, 30 t 31. 32, and 34 of Exercises 13,6.
Figure 13*5?
Review Exercises
611
36. mAB
37. mCT
38. ml BCD
39. nUCPT
40. m£AVC
41. mlCVT
42. mlBTG
43. milBFD
44. mLABD
45. ittilP/V
In Exercises 4650 refer to the circle with center P shown in Figure 1358.
In the figure, VA and VC are secantrays intersecting the circle in points
A* B and C, I ), respectively. VT is a tangent to the circle at 71 Chords AD
and BT intersect at E.
Figure I35S
40, If VA = 12, VB = 20, and VC = 14, find VD and the power of V.
47. If VA = 16, and AB = 9, find VT and the power of V,
48. If AD = 24, AE = 18, and ET m 8, find Brand the power of B.
40, If BE = 12, ET = x, AE = 18, and ED = x  2, find x, BT, AD, and
the power of E.
50. If BE = 9, EA = x, AE = x + 12, and ED = X  3, find *, AE, £D,
and the power of E,
Chapter
Brooks /Mankmeyar
Circumferences
and Areas
of Circles
14.1 INTRODUCTION
In the first part of this chapter, we consider some of the properties
of regular polygons that are useful in developing the formula for the
circumference of a circle and the formula for the area of a circular re
gion. We usually say "the area of a circle" as an abbreviation for the
phrase "the area of a circular region/* or "the area enclosed by a
circle."
Tti the last part of the chapter, we depart from our formal geometry
and present an intuitive approach to the development of the formulas
for the circumference of a circle, the area of a circle, the length of an
arc, and the area of a sector. We appeal to your intuition in developing
these formulas because a formal treatment involves the use of limits,
a topic that vou would study in a mathematics subject called "calcu
lus." We use the idea of a limit to make the formulas seem plausible.
614 Circumferences and Areas of Clrdts Chapter 14
14.2 POLYGONS
In this section we investigate some of the angle measure properties
of convex polygons. We also consider some of the properties of a cer
tain subset of convex polygons called regular polygons.
Recall the definition of a convex polygon in Chapter 4. We say that
a polygon is a convex polygon if and only if each of its sides lies on the
edge of a halfplane which contains all of the polygon except that one
side. In this chapter all the polygons with which we are concerned are
convex polygons. Therefore, when we speak of a polygon, we mean u
convex polygon.
Recall that two vertices of u polygon that are endpoints of the same
aide are called consecutive vertices, or adjacent vertices. Two sides of
a polygon that have a common endpoint are called consecutive sides,
or adjacent sides. An angle determined by two adjacent sides of a poly
gon is called an angle of the polygon. Two angles of a polygon are
called adjacent angles of the polygon if their vertices are adjacent ver
tices of the polygon.
For the polygon ABCDE shown in Figure 141, A and B are ad
jacent vertices. AB and AF, are adjacent sides, and Z.A and LB arc
adjacent angles of the polygon.
Vertices such as A and C, or such as
A and D s or such as B and D* and so
on, are called ncmadjaceiit vertices
of the polygon. A segment whose
endpoints are nonadjacent vertices
of a polygon is called a diagonal of
the polygon. In Figure 141, ~KC is
a diagonal of the polygon. Name
three more diagonals of the polygon
ABCDE. How many distinct diag
onals does this polygon have alto
gether? How many distinct diagonals are there that have a given vertex
as an endpoint ? In the work I hat follows we are going to be concerned
with determining the number of diagonals from an arbitrary vertex of
a given polygon. Note that a triangle has no diagonals since each pair
of vertices in a triangle is a pair of adjacent vertices.
In Chapter 7 we proved that the sum of the measures of the angles
of a triangle is 180. We then used this theorem to prove that the sum
of the measures of the angles of a convex quadrilateral is 360. i See The
orem 7.33 and its proof.) Let us now review the ideas of this proof W \ 1 1 1
the aid of the new terminology introduced in the study of areas in
Chapter 9.
14.2 Polygoni 615
If the quadrilateral ABCD shown in Figure 142 is a convex quad
rilateral, then the diagonal AC parti lions the polygonal region ABCD
into two triangular regions, ABC and ADC. The union of these two
triangular regions is the polygonal region ABCD. In the proof of The
orem 7.33, we showed that the sum of the measures of the four angles
of the quadrilateral is the same as the sum of the. measures of a certain
set of six angles, three from each of the two triangles. In this way we
obtained 2 « 180, or 360, as the sum of the measures of the angles of a
convex quadrilateral.
Now let us extend this idea to convex polygons of 5, 6, 7, 8, or a
sides, where n is a positive integer greater than 4. Figure 143 shows
pictures of polygons with 5, 6, 7, and 8 sides. The names of these poly
gons are pentagon, hexagon, heptagon, and octagon, respectively.
a
616 Circumferences and Areas of Circtat Chapter 14
On a sheet of paper draw four convex polygons, a pentagon, a hex
agon, a heptagon, and an octagon, In each polygon, label one vertex A
and draw all distinct diagonals from A. The diagonals partition each
polygonal region into a certain number of nonoverlapping triangular
regions whose union is the given polygonal region. By a procedure sim
ilar to the one we used with the quadrilateral, find the sum of the meas
ures of the angles of each polygon and summarize the results in a table
like the one shown in Figure 144, Complete the last two rows of the
table on the basis of your computations for the first five rows. The en
tries in the last row shotild l>e formulas involving n.
Number of
Number of
Sum of Measures
Sides of
Number of
Triangular
of the Angles
Convex Polygon
Diagonals from A
Regions
of the Polygon
4
1
2
2 ■ 180 = 360
5
2
m
m
6
3
m
m
7
4
m
m
8
m
m
m
9
m
m
m
n
m
m
Figure 144
On die basis of the results of the computations suggested, it seems
reasonable to conclude that each convex polygonal region of n sides
can be partitioned into n — 2 triangular regions by drawing the diag
onals from one vertex. Since die sum of the measures of the angles of
each triangle diat bounds a triangular region is 180, the formula
(n — 2)180 appears to be a correct formula for determining the sum
of the measures of the angles of a convex polygon of n sides Wc slate
this result as a theorem.
THEOREM 14.1 The sum of the measures of the angles of a con
vex polygon of H sides is (n — 2)180.
Proof: Let A be any vertex of the given convex polygon with n sides
and let the polygon be A BCD . . . MN as suggested in Figure 14*5.
Since a diagonal exists from A to each of the n vertices of the polygon
except the vertices A, B, N (Why?), there are n — 3 diagonals from
the vertex A. Match A ABC with AC, AACD with AD This es
14.2 Polygons 617
tablishcs a onetoone correspondence between the set of n — 3 diago
nals and the set of all triangular regions with the exception of AMN.
Figure 145
Therefore there are (n — 3) + 1, or n — 2, triangular regions. The
union of these n — 2 triangular regions t ABC, ACD, . . . t AMN, is the
polygonal region
ABCD , . . \l\\
The sum of the measures of all the angles of the triangles that bound
these triangular regions is (n — 2)180. It follows from the Angle Meas
ure Addition Theorem that the sum of the measures of all the angles of
the triangles
AAJ5C, AACO t , > . , AAMN
is the same as the sum of the measures of all the angles of the polygon
ABCD . . . MN. This completes the proof.
An important subset of the set of all convex polygons is Lhc set of
polygons whose sides are all congruent and whose angles are all con
gruent. We call such polygons regular polygons. Figure 146 shows a
regular pentagon ABCDE and a regular hexagon ABCDEF.
618 Circumferences and Areas of Circles
Chapter 14
Definition 14 J A regular polygon is a convex polygon ull
of whose sides are congruent and all of whose angles are
congruent.
What do we call a regular polygon of three sides? Of four sides?
Note that a rhombus (Figure 147a) which is not a square has all of its
sides congruent, but it is not a regular polygon. Why? Similarly, a
rectangle (Figure 14~7b) which is not a square has all of its angles con
gruent, but it is not a regular polygon. Why?
Figure 1.47
{a) Rhombus
(b) Rectangle
Since a polygon of n sides has n vertices and therefore « angles,
we have an important corollary of Theorem 14.1 that applies to a regu
lar polygon of n sides.
COROLLARY 14.1,1 The measure of each angle of a regular
polygon of n sides is
(n  2)180
n
Proof: A regular polygon of fl sides has n angles and each of these
singles has the same measure as every other angle of the polygon. Since
the sum of the measures of the angles is (it — 2)180, it follows that each
angle has a measure of  .
n
Tn Chapter 6 we defined an exterior angle of a triangle. We now
extend the definition to convex polygons of more than three sides.
Defin ition 1 4.2 Each angle of a convex polygon is called aj i
interior angle of the polygon. An angle which forms a linear
pair with an interior angle of a convex polygon is called an
exterior angle of the polygon. Each exterior angle is said to be
adjacent to the interior angle with which it Forms a linear
pain
14.2 Polygons 619
Note that there arc two exterior angles at each vertex of a polygon
as suggested in Figure 148 and that, t>eing vertical angles, they are
congruent to each other. It follows that a polygon of n sides has 2n
exterior angles.
Figure 148
Suppose that we are given a convex polygon of n sides and suppose
that we choose one of the two exterior angles at each vertex. The
chosen exterior angle and the adjacent interior angle of the polygon
at that vertex are supplementary. Why? Therefore the sum of their
measures is 180, Since there arc n vertices, the sum of the measures of
dl the interior angles and the chosen exterior angles is n • 180* But we
have shown that the sum of the measures of all the interior angles of
the polygon is (n — 2)180. Therefore the sum of the measures of all
the chosen exterior angles (one at each vertex) is
n(180)  (n  2)180 = 180«  ISOn + 360
= 360.
We have proved the following theorem.
TTTEOREM 14.2 The sum of the measures of the exterior angles,
one at each vertex, of a convex polygon of n sides is 360.
This means that the sum of the measures of the exterior angles of
a polygon is independent of the number of sides the polygon may have.
Suppose we consider a particular regular polygon, such as a regular
hexagon. Wc know that the sum of the measures of all the interior an
gles of the hexagon is (n — 2)180 and that the measure of eacli angle
(n  2)180
it
of a regular hexagon is
For a hexagon, n = 6.
620 circumferences and Areas of Clrclt* Chapter 14
Therefore the measure of each angle of a regular hexagon is
< 6 ~ 2 > 18 ° = 120.
6
It follows that the measure of each exterior angle of a regular hexagon
is 60 and that the sum of the measures of the exterior angles, one at
each vertex, is 6" ■ 60 = 360,
Another way to calculate the measure of each exterior angle of a
regular hexagon is to use Theorem 14.2. Since supplements of con
gruent angles are congruent, It follows that all the exterior angles of a
regular polygon are congruent. If we choose just one exterior angle at
each vertex, there are 6 chosen exterior angles of a regular hexagon
and each of them has a measure of I • 360 = 60, This leads us to the
following corollary of Theorem \42.
COROLTARY 14.2.1 live measure of each exterior angle of a
regular polygon of n sides Is ■ .
Proof: Assigned as an exercise,
EXERCISES 14.2
■ In Exercises 16, use the formula of Corollary i.4.1. 1 to find the measure of
each Interior angle of the indicated regular polygon.
1. Pentagon A. Octagon 5, 24gon (24 sides)
2, Heptagon (7 sides) 4. Decagon (10 sides) 0, ISOgon
7, Fi nd t he in ensure of each exterior angle of the regular polygons of Exer
cises 16 in two differ cut ways,
■ Copy and complete the statements in Exercises 8 and 9 with the word in
creases or the word decreases,
8» As the number of sides of a regular polygon increases, the measure of
each interior angle of the polygon [?J.
&. As the number of sides of a regular polygon increases, the measure of
each exterior angle of the polygon [£}■
10. The measure of each interior angle of a regular ngon is 140. Find n.
{Hint: Solve the equation ( " : i. 2 ) 180 _ 14Q f OT ^j
11. The measure of each interior angle of a regular ngon is 150. Find n.
(See Exercise H>.)
14.2 Polygon* 621
12. Can the measure of each interior angle of a regular polygon be 136?
Explain.
13. Find the measure of each exterior angle of a regular 12gon.
14. The measure of each exterior angle of a regular ngon is 12, Find n.
15. Can the measure of each exterior angle of a regular polygon be 50?
Explain.
16. Given a regular hexagon ABCDEE as shown in the figure below at left,
prove that ABDF is equilateral.
17. The figure above at right shows a regular pentagon ABCDE whose sides
have been extended to form a fivepointed star. Find the measure of
each vertex angle of the star (that is, Z P, LQ> LR? L S, L T).
18. If the skies of a regular hexagon were extended to form a sixpointed
star, what would be the measure of each vertex angle of the star?
19. Hie sum of the measures of 14 angles of a polygon of 15 sides is 2184.
(a) What Is the measure of the remaining angle?
;b) Could the polygon be a regular polygon?
(c) Is there enough information to decide whether it is a regular
polygon?
20. The sum of the measures of 8 angles of a 9gon is 1140.
(a) What is the measure of the remaining angle?
(b) Could the polygon be a regular polygon? Explain.
21. Prove Comllary 14.2.1.
22. Given a regular pentagon ABCDE, prove that diagonal AD is parallel
to side EC,
23. Given a regular hexagon ABCDEl', prove that diagonal AD is parallel
to side BC.
24. challenge PROBLEM^Given a regular polygon ABCD ... .V of n sides,
prove that diagonal AD is parallel to side BC if n > 5.
25. chaij.f.nc;k HKKBJM. Determine the maximum number of acute an
gles a convex polygon can have.
622 Circumferences and Areas of Circle* Chapter 14
14.3 REGULAR POLYGONS AND CIRCLES
We biow that three noneol linear points determine exactly one tri
angle. That is> given a set of three noneollinear points there is exactly
one triangle which has these three given points as its vertices. We now
proceed to show that, given a set of three noncollinear points, there is
exactly one circle that contains the three given points. In this sense,
we say that three noncollinear points determine a circle.
T^t three noncollinear points D, E, F be given as shown in Figure
149, Let of be the unique plane that contains D, E, F. In plane a, let 1%
be the unique perpendicular bisector of DE and let ti be the unique
perpendicular bisector of FE, Let P be die unique point of intersection
of ti and Z 2  (How do wc know that k intersects Z 2 ?) Then P is equidis
tant from D and E because it lies on the perpendicular bisector of IM>
and P is equidistant from E arid F because it lies on the perpendicular
bisector of EF. Therefore P is equidistant from D, E, and F. It follows
that P is the center of a circle C with radius
r = PD = PE = PF.
hf.^
Figure 149
Therefore C is a circle which contains D i E x F* Furthermore, C is the
only circle which contains D t £, F. For if C were any other circle with
center F and radius r* and containing points D, F., F in plane a, then
¥ = FD = FE=FF
and F would be equidistant from D, E, and F, Therefore F would lie
on h and 1% the perpendicular bisectors of DF and EF in plane a. Since
there is only one point that lies on both of these lines, F = P> ¥ = r,
and hence C = C. Therefore there is exactly one circle which contains
any given set of three noncollinear points. It follows that there is
exactly one circle that contains the vertices of any given triangle,
We have proved Theorem 14.3.
14.3 Regular Polygons and Circles 623
Definition 14,3 The circle which contains the three vertices
of a given triangle is called the circumscribed circle, or cir
cumcircle, of the triangle arid we say that it circumscribes
the triangle. The triangle is said to be inscribed in the circle
and is called an inscribed triangle of the circle.
THEOREM 14.3 A given triangle has exactly one dreumeirde.
We now extend the statement of Theorem 14.3 to Include any reg
ular polygon, Wc want to prove that, given any regular polygon, there
is exactly one circle that contains all the vertices of the polygon (that
is. the given regular polygon has exactly one circumcircle).
THEOREM 14.4 A given regular polygon has exactly one
circumcircle.
Proof: Let a regular polygon of n
sides be given. (In Figure 1410, we
have shown a polygon of 6* sides.) Let
Q be the unique circle with center P
and radius r which contains A, B,
and (7. (How do we know this circle
exists and is unique?) We shall prove
that PD = r and hence that D lies on
circle Q, A similar argument could be
given to show that each of the verti
ces of the given polygon lies on Q,
Figure 14>10
1. PA = PB = PC = r
2. Z2s Z3
3. mLABC = m/BCD
4. mZ2 = m/.$
5. mZl = mZ4
6. Z 1 == 14
7. AB S DC
8. PB m PC
9. &PBAj= APCD
10, PA B PD
11. PD= PA = r
Reason
1. Definition of circumcircle
2. Why?
3. Why?
4. Why?
5. Angle Measure Addition
Theorem (3, 4)
6. Why?
7. Why?
8. Why?
9. S.A.S, Postulate {7, 6, 8)
10, Why?
11. Statements JO and 1
Thus D is on Q and this completes the proof for point D,
624
Circumf»r»nc«s *nd Artai of Clrcltt
Chapter 14
Definition 14.4 {See Figure 1411.) The rlmrffimfWllir of a
regular polygon is the center of its circumscribed circle. A
circumradius of a regular polygon is a segment (or its length)
Joining the center of the polygon to one of the vertices of the
polygon. An inradius of a regular polygon is a segment ior its
length) whose endpoints are the center of the polygon and
the foot of the perpendicular from the center of the polygon
to a side of the polygon, A central angle of a regular polygon
is an angjle whose vertex is at the center of the polygon and
whose sides contain adjacent vertices of the polygon.
figure I Ml
Some authors use radius Instead of circumradius, and apotfiem
instead of in radius. We prefer the more descriptive terms* circumradius
and hirudins. They are the radius of the circumscribed circle and the
radius of die inscril>ed circle of a regular polygon. We know that a
regular polygon has a circumscribed circle (Theorem 14.4) and we shall
see that a regular polygon has an inscritwd circle (Theorem 14.7).
Note that, in connection with a regular polygon, each of the words
circumradius and inradius is used in two different ways: ( 1 ) as one of a
set of segments and (2) as a positive numl>er. For example, the inradius
of a regular polygon means the number that is the length of a segment
— any one of the segments defined as an inradius in Definition 14.4.
We shall sec in Theorem 14.8 that all these segments have the same
length. On the other hand, an inradius of a regular polygon is a seg
ment, usually one of the segments defined as an inradius in Definition
114, but it might also mean any radius of its inscribed circle. Similarly,
the circumradius of a regular polygon means a number, whereas a
circumradius of a regular polygon usually means any one of the seg
ments joining the center of the polygon to a vertex of the polygon, but
it might also refer to any radius of die circuinscril>ed circle. The con
text in which the word is used should make it easy to decide which
meaning is intended.
It follows from the definition of a central angle that a regular poly
gon of n sides has n central angles.
14.3 Regular Polygons and Circles 625
Suppose that we are given a regular polygon ABCD . . . MN of n
sides. (Figure 1412 shows a regular polygon of 6 sides.) Let F be the
circunicentcr of the given polygon. Il follows that all of the triangles,
APAB, AFBC, A PCD, , . , , APJVA, are congruent by the S.S.S.
Postulate.
Kipire 1412
(Prove that A PAS = A PUG) Let us agree to call each of these tri
angles (that is, a triangle whose vertices are the center and the end
points of a side of the polygon) a central triangle of the regular poly
gon. It follows from the definition of congruent triangles that all of the
central angles of a given regular polygon are congruent. We combine
these results into the statement of our next theorem.
TI7EOREM 14.5 Let a regular polygon of n sides be given. Then
all the central triangles of the gjven polygon are congruent and all
of the central angles of the given polygon are congruent.
Again, suppose that we are given a regular polygon ABCD . . . MN
of n sides. (Figure 1413 shows a regular polygon of 6 sides.)
Figure 1413
Let P be die cii'cumcenter of the given polygon and let PR, PS, FT, » . « »
PVbe the n inradii of the polygon. By the definition of an inradius, the
626 Circumferences and Area* 0* Circle* Chapter 14
segments PR, M, PT, . . . , PV are the altitudes to the bases IB, BC, CD,
. « < , RA, respectively, of the central triangles, APAB, APBC, APCD,
. , . , APNA, of tlic given polygon, By Theorem 14.5 these central tri
angles are all congruent. Since corresponding altitudes of congruent
triangles arc congruent, it follows that all the inradii of a given regular
polygon are congruent This means that, in the plane of the given poly
gon, the points R, $, T Vile on a circle {) whose center is P and
whose radius, a, is the inradius (a number) of the polygon. Since
PR I AB^PS^ SC^ff A OR . . . t FT A KE, it follows that each of
the sides AB, BC, CD* . . , , NA of the given regular polygon is tangent
to the circle Q and that the points R, S, T, , V are their respective
points of tangeney.
Definition 14.5 A circle is said to be inscribed in a polygon
and is called an inscribed circle, or iiicircle, of the polygon if
each of the sides of the polygon is tangent to the circle. We
also say that the polygon circumscribes the circle. The center
of an incircle of a polygon is an incenler of the polygon*
We now show that there is exactly one circle inscribed in a given
regular polygon. Let a regular polygon such as the one in Figure 1413
be given.
We have demonstrated that the circle Q with center P and radius a
is an inscribed circle of the given polygon. It can be shown that a point
which is equidistant from the sides of a regular polygon is also equi
distant from its vertices and hence must be the circumeenter. Since
there is only one circumeenter, it follows that there is only nine ineen
ter. Therefore Q is the only inscribed circle of the polygon. We have
proved the following two theorems.
THEOREM 14.6 All the inradii of a given regular polygon are
congruent
THEOREM 14.7 There is exactly one circle that is inscribed in
a given regular polygon.
We can now think of the center of a regular polygon as its incenter
or its circumcenter.
Recall from Chapter 10 that two polygons are similar if there is a
correspondence between their vertices such that corresponding angles
are congruent and lengths of corresponding sides are proportional. It
14.3 FUgular Polygons and Circles 627
follows from the definition of similar polygons that if two regular
polygons are similar, then they have the same number of sides, Is the
converse statement true? That is, is it true diat if two regular polygons
have the same number of skies, then they are similar? This brings ns
to our next theorem,
THEORFM 14,8 Two regular polygons are similar if they have
the same number of sides.
Proof: Let two regular polygons ABCD , . , M .V and
A'B'C'D' , . » M'N\ each having n sides, be given as suggested in Fig
Figure 1414
ure 1414 We want to prove that the correspondence
ABCD . . . MN * — * A'B'Ciy . . . 1TW
is a similarity. By the definition of regular polygon and Corollary 14. LI,
we have
m
n
and
mLA* s m£W = m£C = mlD' = • • • = mlN'
_ (n  2)180
n
Therefore
mLA = m£A\m£Sm mlB' mlN = mlN',
and the corresponding angles of the two polygons are congruent.
628 Circumferences and Areas of Circles Chapter 14
It follows from the definition of regular polygon that there are two
positive numbers 9 and $' such that
AB = BCCD = •> =NA = s
and
AW a B'C = CD' m , , . = N'A' = s'.
Then
AB = s = £■•• = 4A'F, BC = * = 4**' ■ 4B^ etc.
It follows that
(AB, BC. CD, , . , , NA) = (A'B\ B'C, CD\ , . ♦ , iVA )
with proportionality constant k = ^ , Therefore the lengths of the cor
responding sides of the two given polygons are proportional and
ABCD ...MN~ A'&Cfy .  . M'N'.
This completes the proof of Theorem 14,8.
THEOREM 14.9 The perimeters of two regular polygons, with
the same numl>er of sides, are proportional to the lengths of their
circiunradii or their in radii.
Proof: Let two regular polygons ABCD . . . MN and A'B'Ciy . . .
M f N* t each having n sides, be given as suggested in Figure 1415.
A H B
Figure 1415
Let p and p', s and $', t and r\ a and a' be their respective perimeters.,
lengths of sides, circumradii, and inradii. Let P and W be the centers of
14,3 Regular Polygons and Circles 629
the two polygons and consider the two central triangles
AAPB of ARCD . . , MN
at id
AA'FB' of A'B'CD' , , . Af.V.
Since
ml ABB = mlATB' =
n
(you will be asked to show this in the Exercises), it follows that
AAFB — AA'FB' by the S.A.S. Similarity Theorem, Therefore
(AB.r,a) = (A r B' f r>,tri. Why?
Since corresponding altitudes of similar triangles are proportional to
the lengths of any two corresponding sides, we have
($, r, a) = {s\ f, a'}.
Finally,
p = ns for ABCD . . , MX and p' = ns' for A'B'CD' . . . M'.Y.
Since
ps' = (ns)s' = (ns> = pfs,
it follows that
and hence
(p. s, r» a) =: (p\ *', /, a').
This completes the proof of Theorem 14.9*
Since all central triangles of a given regular polygon of n sides are
congruent, and since they partition the given polygonal region Into n
nonoverlapping triangular regions, it is easy to prove that the area of
a regular polygon is equal to onehalf the product of its inradius and
perimeter. We state this as our next theorem,
TITEORFM 14 JO The area of a regular polygon is equal to one
half the product of its inradius and perimeter, that is, S = jflp.
Froof: Assigned as an exercise.
630 Circumference* and Aft*« vt Clrclei Chapter 14
The last theorem of this section follows easily from Theorems 14.9
and 1410,
THEOREM 14,11 The ureas of two regular polygons with the
same number of sides are proportional to the squares of their
circum radii (or the squares of their in radii, or the squares of their
side lengths, or the squares of their perimeters).
Proof: Assigned as an exercise.
As we said in the introduction to this chapter, the idea of a limit is
important in developing formulas for the circumference and the area
of a circle. We conclude this section with a brief discussion of se
quences and limits.
An infinite sequence of numbers, denoted by {%„}, is a sequence
*i» *fc *& • • • , *■» • * « in which, for every positive integer n, *„ is a
number. In some applications it is convenient to start counting with
some integer other than 1 . Thus £3, jt, *a, . . . , x n> . , . is an infinite
sequence.
Example 1 Let the nth term of a sequence be given by x„ = 2n.
Write the first five terms and the 40th term of the sequence {2n},
Solution;
*! = 2 • 1 = 2.
*a = 2*2 = 4.
* 3 == 2*3 = 6,
x, = 2 ♦ 4 = 8.
* 5 = 25 = 10.
xw = 2 ■ 40 = 80.
Therefore die first five terms of the sequence are 2, 4, 6, 8, 10 and the
40th term is 80,
Example 2 Let the nth term of a sequence be given by
1* + 1
*» =
n
Write the first five terms, the J 00th term, and the 1000th term of the
sequence
m
Solution:
1
+
1
*1 =
1
—
2.
*2 =
2
+
2
1
—
3
2
*3 =
3
+
1
4
3
3
4
+
1
5
X4 =
4
=
4
14,3 Regular Polygons and Circles 631
5 + 1 6
*S = = — = ~ *
5 5
IftA . 100+1 101
100 " 100 " 100
1000 + 1 1001
* ,000= 1ooo~iooo
In Example 2, as n gets larger and larger, the terms of the sequence
gj$f closer and closer to some particular number. What number is it?
In Example 1, as n gets larger and larger, do the terms of the
sequence {2n} get closer and closer to some particular number? If
so, what number is it?
If the terms of a sequence {x n } get arbitrarily close (as close as we
desire) to a particular number L as n gets larger and larger, we then
say that the nth term of the sequence is approaching L (denoted by
x n —*L) and we call L the limit of the sequence. Thus, in Example 2 S
n
and we call 1 the limit of the sequence  1 , In Example 1, the
sequence {2n} has no limit Find the limit (if any) of the sequence — 1 .
EXERCISES 14.3
L It follows from Definition 14.4 that a regular polygon of n sides has n
central angles. Prove that the measure of each central angle of a regular
360
polygon is — — .
2. Prove that each central triangle of a regular polygon is an isosceles
triangle.
3. Prove that each central triangle of a regular hexagon is an equilateral
triangle.
4. Prove that the length of a side of a regular hexagon is equal to the
circumradius of the circuiucirclc.
632 Circumferences and Area* of Circlt* Chapter 14
5. Prove thai an in radius of a regular polygon bisects a side of the polygon
and hence lies on the perpendicular bisector (in the plane of the poly
gon) of the side of the polygon,
6. Prove that the bisector rays of the interior angles of a regular polygon
are concurrent at the center of the polygon.
7. Prove that the measure of a central angle of a regular polygon is equal
to the measure of an exterior angle of Lhe polygon,
8. Two regular pentagons have sides of length 4 in. and 5 in,,, respectively .
What is the ratio of the perimeter of the smaller pentagon to the
perimeter of the larger one? What is the ratio of their eircurnradii? Of
their inradii? Of their areas?
9. The area of a regular 12gon is *£■ times the area of a second regular
12gon. What is the ratio of their perimeters'? Of their eircuinradii? Of
their inradii?
10. A regular hexagon has twice the area of another regular hexagon. What
is the ratio of the perimeter of the smaller hexagon to the perimeter of
the larger one? What is the ratio of the lengths of their sides? Of their
cirenmradii?
11. Two regular polygons of the same number of sides have perimeters of
36 in. and 48 in., respectively. The in radius of the first polygon is
3\/5hx What is the hirudin* of the second polygon? What is the area
of each of the polygons?
12. Find the circum radius, inradius, and area of an equilateral triangle
each of whose sides is of length 2 \/3.
13. Find the circum radius, hirudins, and area of a regular hexagon each of
whose sides is of length 12.
14. Show that the inradhis of a regular hexagon is — ^ — . s, where s is the
length of a side of the hexagon,
15. Derive a formula for the urea S of a regular hexagon in terms of the
length s of its side. (Hint: See Exercise 14.)
16. Use the formula you derived in Exercise 15 to find the area of a regular
hexagon each of whose sides is of length 12. Does your answer for die
area agree with that of Exercise 13?
17. If a regular hexagon and a regular triangle are inscribed in the same
circle, prove that tire length of the side of the Hexagon is twice Lhe ui
radius of the triangle,
18. A square is inscribed in a circle of radius 1, A second square is circum
scribed about the same circle. Find the area and the perimeter of each
square,
19. Repeat Exercise 18 using regular hexagons instead of squares.
20. The length of each side erf a regular hexagon is 8 s/3. Find the area of
the hexagon in two different ways.
14,3 Regular Polygon* and Circfct 633
21* The circum radius of a regular pentagon is r and the length of each of
its sides is s. Find the area of the pentagon in terms of r and s.
22. In Exercise 21, if the eircumradius of a second regular pentagon is 2r,
find die area of the second pentagon in terms of t and *.
23. Prove Theorem 14 JO,
24. Prove Theorem 14 J L
25. Write reasons for statements (2), (3), (4), (6), (7), (8), and (10) In the
proof of Theorem 14,4.
26. What is the 1000th term of the sequence {}? What is the millionth
term? Which of these two terms is closer to zero? Could we make — as
n
dose to zero as we desire by choosing n large enough?
27. Find the first five terms, the 100th term, and the 1000th term of the
28. What is the millionth term of the sequence 1 ^ 1 of Exercise 27?
Find (he limit (if any} of the sequence.
In Exercises 29^37, a formula for x m is given. Find *i, x», x& and Xi<>, What
is the limit (if any) of the sequence («■}?
■fc^l + i M.^iiJOO 1 + 50
~2n 2u 2 tt
2n n
* L ** + 5T 2n m *2nT
32.^=2 y? tXm = !£jLL
33, ^ = J
2"
38. Let a circle be given. For each n,n > 3, let a regular polygon of n sides
be inscribed in the circle. Let »„ be the perimeter of the ngon. Does
Px, p<» Ps» • ♦ ■ , p«» ■ • • define a sequence? Do you think {p„ } has a limit?
If so, how would you describe the limit?
39. 1 ,et a circle be given. For each n, n > 3, let a regular polygon of n sides
be inscribed in the circle. I .et S n be the area of the polygon. Does S3. S*.
Sa, . , , , S nw . , . define a sequence? Do you think {S„} has a limit? If so,
how would yon describe the limit
"J^ . — jT — J has a limit.
What is it?
634 Circumferences and Are« of Circle*
Chapter 14
14.4 THE CIRCUMFERENCE OF A CIRCLE
Thus far, in our formal geometry, length has not been defined for
anything except segments. If a path from one point to a second point is
such that every point of the path lies on the same segment, then the
length of that path is, of course, the length of the segment joining the
two points. However, If the path is a circular arc, what is the distance
from the first point to the second point along the circular arc, that is,
what is the length of the arc? The degree measure of the arc would
not be a satisfactory way of describing its length since it is possible
for two arcs to have the same degree measure and Lo have different
lengths as suggested in Figure 1416.
Figure MI6
Each of die arcs AB and A B' shown in the figure has a degree
measure of 90. But it certainly seems reasonable to think of the arc AB
as having a greater length than the arc A B\ Wc start by explaining
what we mean by the length of circular arcs and then deriving ways
of finding such lengths. We first proceed informally, referring to the
physical world.
You may have been asked, at one time or another in yonr study of
informal geometry, to wrap a string around a circular object, then pull
it out straight, and measure its length, liy doing this you are able to
arrive at an approximation to the distance around the object However,
we cannot describe the process of wrapping a string around a circular
object in our formal geometry.
We call the '"distance around a circle" the circumference of the
circle and denote it by C A more sophisticated approach to finding an
approximation to the circumference of a dido is in terms of the perim
eLers of regular polygons inscribed in the circle. There was no difficulty
14.4 Th« arcumftmtc* of • Circle 635
in defining the perimeter of a polygon because the sides of a polygon
are segments and each of these segments has a length. But a circle con
tains no segment of a line, and thus we cannot define its circumference
(or perimeter) so simply. It seems reasonable lo suppose that if we want
to find the circumference of a circle approximately, we can do it by
inscribing in the circle a regular polygon with a large number of sides
and then measuring or computing the perimeter of the polygon*
Given a circle, let »„ be the perimeter of a regular polygon of n sides
inscribed in the circle. Then as n gets larger and larger, the number p„
increases, that is, each term in the sequence {p„} ts greater than the
preceding term. For example, we can inscribe a square in a circle. By
bisecting the central angles of the square we obtain a regular octagon
inscribed in the same circle as shown in Figure 1417. Using the Tri
angle Inequality Theorem, it is easy to show that the perimeter p$ of
the regular octagon is greater than the perimeter p 4 of the square. If we
continue to bisect the central angles, wc obtain a regular 16gon, a reg
ular 32gon, and so on. The perimeters of these regular polygons p 4 . ptt*
Pig* P32. .  ■ form an infinite sequence of numbers, and each term in
the sequence is greater than the preceding term.
Hfttra 1417
It is shown in the calculus that if a sequence of numl>ers is increas
ing (that is, if each term in the sequence is greater than the preceding
term), and if the sequence is bounded (that is, if there is a number that
is equal to or greater than any term in the sequence)* then the sequence
has a limit. The squcnoe p*» pg, pm, ps*jt • ■ ■ described above in con
nection with the circle shown in Figure 1417 can be shown to be
bounded. In fact, it can be shown that any square that circumscribes
the given circle has a perimeter that is greater than any of the terms in
the sequence.
636 Cfreumfcrancet and Areas of Circlai Chapter 14
Let a sequence of perimeters of regular polygons of n sides in
scribed in a given circle be denoted by {p«}. Let die limit of this se
quence be denoted by C, that is, Hm p n = C. We are now ready for
our formal definition of circumference as die limit of the p, .
Definition 14.6 The circumference of a circle is the limit
of the sequence of perimeters p a of the inscribed regular pol
ygons (that is, C = lim p n ).
Note that we are forced to use limits in defining the circumference
of a circle. In order to derive the formulas for the circumference and
for the area of a circle* we need a theorem about limits which we state
here without proof. We can treat the theorem as a postulate, although
it is not a postulate concerning our formal geometry. Rather t it is a
postulate concerning the real number system.
THEOREM 14.12 Let {**} and {y n } be two sequences of real
munhers with nth terms x n and y n , respectively,
1 . If the limit of x« is Lj and the limit of y n is I^ t then the sequence
whose nth term is x„y n has a limit, and
ton x n y n = Li'Lz.
2. If die limit of *„ is Lj and the limit of y H is L^ J* 0, then the
sequence whose nth term is — has a limit and lim — = — .
3. If x n = y„ for every positive integer n > I and if {*„) and { y„]
each has a Limit, then lim x» = lim i/„.
4 If k is a real number and if x n = k t for every n > l f then
lim x n — k.
Example 1 If lim ^^=A = 2 and lim ^ = 3, then
n n + 2
rnl
U,n(fczLa._aL) = 23 = 8
\ n n + 2/
\ n n f 2/ 3
2 Find lim x n if x?, = 2 for every » > 1.
14.4 Tht Cfrcumferancs of a Circle 637
Solution; If x, { = 2 tor every n > 1, then {ar rt } is a sequence of num
bers whose every term is 2; that is, {a&,} = 2, 2, 2 By part 4 of
Theorem 14.12, lim x n = 2,
Before we derive a formula for the circumference of a circle, we
need to know that the number ~ , where C is the circumference of a
a
circle and d is its diameter, is the same for all circles, That the num
ber ~ is the same for all circles is a corollary of our next theorem.
THEOREM 14.13 If C and C arc the circumferences of any two
circles with diameters d and il\ respectively, then
(C, d) = (C. <f).
Proof: Let K t K' be any two circles with circumferences C. C and
radii r, r', respectively, as shown in Figure 1418.
Flf»UIfi
Let [p H ] he the sequence of perimeters of regular polygons of n
sides inscribed in circle K with radius r and let [p^] be the sequence of
perimeters of regular polygons of n sides inscribed in circle K* with
radius r\ By Theorem 14.9,
(p», r> f (pi />.
It follows that
hence that
(1) (Pm <*) = (Pn, <#%
where d and ci are the diameters of the circles K and K', respectively.
638 Circumferences and Artai of Ckrd«i ChapUr 14
By Definition 14.6,
lim p n = C
and
Ump^C,
It follows from Equation (I) that
(2) £L = iL.
By part 3 of Theorem 14.12, we get from Equation (2)
(3) lim &■ = lim 4
By part 1 of Theorem 14.12,
lim £=■ = £,
and, since d and d' are the same for all n, it follows from part 4 of The
orem 14.12 that
hm = .
Substituting these last two results in Equation (3), we have that
SL±
C d'
or that
and the proof is complete.
COROLLARY 14.13.1 If C and d are the circumference and di
ameter, respectively, of a circle, then the number ^ is the same for
a
all circles.
Proof: Let K, K' be any two circles with circumferences C, C and
diameters d, d\ respectively, By Theorem 14.13, we know that
(C, d)  «7, d'\
14.4 The Circumference of a Circle 639
By alternation, we get
(C, C) = (d, d').
Therefore y = ■*£, This proves that the number r is the Same for any
two circles.
Definition 14.7 If C is the ctreum fcrence of a circle and d is
its diameter, then the number 4, which is the same for all
a
circles, is denoted by the Greek letter r.
It follows from Corollary 14,131 and Definition 14,7 that
C = ml
for any circle with circumference C and diameter d. Since d = 2r,
where r is the radius of the circle with diameter d t we have
C = 2*rr
as another formula for the circumference of a circle.
It has been proved that it is not a rational number, that is, v cannot
be represented by ^ where a and b are integers with b =fi 0. However,
h
we can approximate tt as closely as we desire by means of rational num
bers. Some of the more common rational number approximations to sr
are 3.14, 2 ^, and 3.1 416. It has been shown that ff, to ten decimal
places, is 3.1415926535.
EXERCISES 14.4
1. Show that f{ is a closer approximation to w than is 2 ? 2
2. Show that 2 f z: 3,14 to the nearest hundredth. (We read "^" as *'is
approximately equal to.")
In Exercises 310, C, r t and d represent the circumference, radius, and di
ameter, respectively, of a given circle. Express answers in exact form, in
terms of it if necessary.
3. If r = 7, find a
4. If C = 83«r, find ft
5. If d =12.5, find C.
640 Circumferences and Ar»n of Circle* Chapter 14
6. If f = 36.4 in., find C.
7. If C = 36.4 in., find r.
& If C = 2**, find r.
9, If r = 3ir, find C.
10. If C= 14*26*, find Id
1L In Exercise 6, use v = 4^ and find C to tlie nearest inch.
12* In Kxerdse 6, use % = 3.14 and find C to the nearest inch. Compare
your answer with that of Exercise 11.
13. In Exercise 7, use r, = 3^ and find r to the nearest hundredth of an
inch.
14. In Exercise 7, use *r = 3.14 and find r to the nearest hundredth of an
inch. Compare your answer with that of Exercise 13.
15. Prove that the circumferences of two circles are proportional to their
radii,
10. The circumference of one circle b twothirds the circumference of a
second circle. What is the ratio of the radius of Oie first circle to the
radius of the second circle?
17. Two circles have radii of 15 and 23, What is the ratio of the circum
ference of the smaller circle to the circumference of the larger deck?
What is the ratio of the diameters of die two circles?
18. A lire on a wheel of a car has a diameter of 28 in. 1 f the wheel makes 12
revolutions per second, wliat is the approximate speed of the car in
miles per hour? (Use w == 4j^.)
19. Given the same car as in Exercise 18, how many revolutions per second
would the wheel make if the car were traveling 50 miles per hour? (Use
tr= 2 7 2 )
20. A square ABCD fa inscribed in a circle with center E as shown in the
figure. If the radius of the circle is 7, find the perimeter of the square.
(Hint: Show that &EAB is an isosceles right triangle.)
14.4 The Circumference of a Circle 641
21. A square ABCD is inscribed in a circle with center £, Another square
PQHS is circumscribed about the same circle as shown in the figure. If
the radius of the given circle is 14 in., find the following:
(a) The perimeter of each square to the nearest inch.
(b) The area of each square to the nearest square inch,
(c) The circumference of the circle to the nearest inch. (Use v = Q)
v
E
B
22. The perimeter of a square is equal to the circumference of a circle. If
the radius of the circle is 1, find the area of the square in terms of ?■
23. Show that if the radius of a circle is increased by 1 unit, the circumfer
ence of the circle is increased by 2*r units.
24. Show that if the circumference of a circle is increased by 1 unit, the
radius of the circle is increased by «=r« units.
7 2ff
25. Assume that the surface of the earth is u sphere. Imagine that a steel
band is fit snugly around iho equator (a great circle of the earth";. Sup
pose that one foot is added to the length of the band and that it is raised
a uniform amount all the way around the earth. To the nearest inch.
how many inches will the new band be above the earth? (See Exercise
24.)
26. Assume that the surface of an orange is a sphere. Imagine that a steel
band is placed around a great circle of the orange so that it just fits. Sup
pose that one foot is added to the length of the hui id and that it is raised
a uniform amount all the way around the orange. To (lie nearest inch,
how many inches will the new bund Ixs above tlic orange? (See Exer
cises 24 and 25.)
642 Circumference* and Ar»« of C4rol»»
Chapter 14
Exercises 2736 refer to Figure 1419. ABCD is a square inscrilwd in the
circle with center P, The bisector rays of the four central angles of square
ABCD intersect the circle in points E t h\ G» H t respectively. The polygon
AEBFCCDH is a regular octagon inscribed in the circle, and TQRS is a
square circumscribed about the circle with points of tangcucy E, F, C, H.
The radius of the circle is 8. If an answer to an exercise is an irrational nunt
her, give a rational approximation to the nearest tenth.
Figure l+iu
27. Find the length * of a side of square ABCD,
28. Find the perimeter of square ABCD.
29. Find the area of square ABCD,
30. Find the length ** of a side of the regular octagon AEBFCCDH.
31. Find the perimeter of the octagon.
32. Find the area of the octagon.
33. Find the length s" of a side of the square TQRS.
M. Find the jwrimeter of square TQRS.
35. Find the area of square TQRS.
36. Find the circumference of the circle.
37. Imagine an infinite number of regular polygons inscribed in a circle
with radius r. The first polygon has 3 sides, die second polygon has
4 sides, the third has 5 sides, and so on. For every n, n > 3. let p„, a n , ft*
and S, he the perimeter, iriradius. side length, and area, respectively,
of the regular inscribed polygon with n sides. What is lim p m ? lim a£
I in i *»? lim S m ?
14.5 Area* of Circle*. Arc Length: 5ectw of a Circle 643
14.5 AREAS OF CIRCLES; ARC LENGTH; SECTOR OF A CIRCLE
In Chapter 9 we considered areas of polygonal regions. Recall that
a polygonal region is the union of a polygon and its interior. In this sec
tion we are concerned with areas of circular regions. We make the
following definition.
Definition 14.8 A circular region is the union of a circle and
its interior.
As we said at the beginning of the chapter, "the area of a circle" is
m abbreviation for "the area of a circular region," or for "the area
enclosed by a circle."
We now proceed to get a formula for the area of a circle. We
already have a formula for the area of a regular polygon of n sides
which is
where a n is the inradius of the polygon, p„ is the perimeter of the
polygon, and S„ is the area of the polygon.
If P„ is a regular polygon of n sides inscritied in a circle with center
Q and radius r t as shown in Figure 1420 (in the figure, n = 8), we
observe that the area of the inscribed polygon is less than the area of
the circle.
Figure 1420
644 Circumf irenc** and Aran of Clrcltt Chapter 14
For the expression
where n = 3, 4, 5 t . . . , there are three sequences involved;
{S n }, {a n ) t and {p n }.
t*et us consider each of these sequences separately,
1. The sequence {*>„}. As noted above, for each n, S„ is always less
than S, the area of the circle, The difference between S s and _S
can be made arbitrarily small by talcing n large enougli. It seems
reasonable then to say that
lim S» = S.
Definilon 14.9 The urea of a circle is the limit of the se
quence of areas of the inscribed regular polygons.
Thus lim S„ = S, by definition.
2. The sequence [a„). Since the length of a leg of a right triangle
is less than the length of the hypotenuse of the right triangle, we
observe that the inradfus <*, is always less than the radius r of
the circle for each particular value of n, However, the difference
between a H and r can be made arbitrarily small by choosing n
laige enough Thus it seems reasonable to say that
lim On = r,
and we accept this feet without proof.
3. The sequence {«„}, By the definition of the circumference C
of a circle,
lim p n = C,
Now we have
S n = fop*.
By parts 3 and 4 of Theorem 14.12, we get
(1) lim S„ = ^lim a n p n >
14.5 Atms of gmlM, Are Ungth; Sector of a Circle 645
By definition 14.9,
2 : lira S» = S.
By part J of Theorem 14.12, we get
(3) lim anpn = lim o„ ■ lim p n
But
(4) lim (in = r
and
(5) lim p rt = C.
Substituting the results of (2), (3), (4), and (5) in Equation (1),
we obtain
(6) S = {tC
as a formula for the area of a circle. Since
C = 2*rr,
we get by substitution into Equation (6),
S = ^r2irr
or
S =
as a formula for the area of a circle— a formula that should be
familiar to all of you.
We now state this result formally as a theorem.
THEOREM 14.14 The area S of a circle with radius r is wr 2 , that
is,
COROLLAR Y 14.14, 1 The areas of two circles are proportional
to the squares of their radii.
Proof; Assigned as an exercise.
We have defined the circumference of a circle to be the limit of the
sequence of the perimeters of the inscribed regular polygons. We now
proceed to define the length of an arc of a circle as a certain limit.
646 Circumftnncti and Areas of Circfti Chapter 14
Consider an arc AM of a circle with center V as shown in Figure
1421. For k > 2, let P u Pa, »%..., Phl be points on Ait such that
each of the & angles
ZAVP b ZP i VP 2 , / P*VP 3 , . .. , Z ft., VB
has a measure of ~ m^C3. (in Figure 1421, k = 4.)
Figure 1421
Let
A* = AP t + PjJPa + f»p, + . . . + P*iR
Tiros, in Figure 1421,
A 4 = APi + Pjft + P^ + P 3 B.
If we bisect each of the central angles
£AVP U L PiVP 2 LPhiVB,
we obtain k more points Q u fy Q k onAB such their
A 2 * = AQ t + QiP x + P^ 2 f Q, £ p 2 + . . . + Q fc R
In Figure 1421,
As = (A0, + Q l P 1 ) + (P^g + £*P 2 ) + (P 2 £ 3 + ^) 3 p 3 )
+ (p*q 4 + g^.
Now
APi + Pi^i > APi,
F,p 2 + <?aP 2 > PiPa,
Pap3 4 Ws > PaP*
^Q* + Q 4 B > P3H, Why?
14.5 Arras of Circles; Arc Langth; Sector of • CJrcl* 647
It follows that As > A*. If we continue to bisect the central angles
with vertex V, we obtain a sequence of sums
A*, As, Aje, . . •» A„
(where n = k, 2k t 4k t &k t ,* .) in which each term in the sequence is
greater than the preceding term. Also, this is a bounded sequence.
Each number in it is less than the circumference of the circle. There
fore the sequence {A n } has a limit and we define that limit to be the
length of arc A B,
Definition MAO The length of arc AB (denoted by I A B) is
the limit of [A m ] where
A„ == APi + P X P 2 + • •  + PniB
and where Pi, P> £ F n i are n  1 distinct points of A&
subtending congruent angles at the center V of the circle
containing AH.
We now have two types of measure for arcs of a circle: their de
gree measures and their lengths.
Definition 14 J I In the same circle or in congruent circles,
two arcs arc congruent if and only if they have the same
length.
Thus, if arcs AB and &W are arcs of congruent circles and if
AB 5b A'B\ we have
(1) IA3 = IA T &'*
Also, from Definition 1347, if AB s X" # B\ we have
(2) mAB = mA%.
Hence, if A& and A^T are congruent arcs of congnient circles, then
(JAB, life) = {mAB, mAW).
lu other words, lengths of congruent arcs of congruent circles are pro
portional to their degree measures. This is a trivial assertion, of course.
Suppose, for example, that IAB = 100 and mA$ = 20. What we are
inserting is that
(100, 100) = (20, 20).
648 Circumferences and Ar*i« o» Circtoi
Chapter 14
It is also true (not trivial and \vc shall not prove it here) that the lengths
of any two arcs of congruent circles are proportional to their degree
measures. We state this as our next theorem.
THEOREM 14,15 The lengths of arc* of congruent circles are
proportional to their degree measures.
Thus, if K and K' {as shown in Figure 1422) are congruent circles
and if A& is an arc of K and A'B' is an arc of K\ we have
(IAB> U%) = (mAB t mAW).
Suppose that A'B' in Figure 1422 is a semicircle. Then
niAW = 180 and LYB' = *r. Why?
Let b\B 1ms denoted by L and mAB be denoted by M. It follows from
Theorem 14.15 that
Therefore
and
(/,, vr) = (M, 180).
180L = tttM
We have proved the following theorem,
THEOREM 14,16 The length L of an arc of degree measure M
contained in a circle with radius r is uiLWr, that is,
\ 1 g!U/
I
\180/
14.5 Area* of Circle*; Arc Untfh; Sactw of a Circle W9
Example 1 Find the length of an arc of a circle with radius 12 if the
degree measure of the arc is 60.
Solution; Use the formula of Theorem 14.16 with M = 80 and
f = 12, Therefore
Exam pit 2 Find the degree measure of an are of a circle with radius
2 if the length of the arc is 4? ■
•i
Solution: Solving the formula given in Theorem 14,16 for M, we
M _ (J80U (Show this.)
to
3
We are given that L = 4? and that r = 2. Therefore
MJ3L& 120.
sr«2 3
Note that if M = 360 in the formula of Theorem 14.16, we get
Thus L equals the circumference of the circle as it should.
Figure 1423 shows a portion R of a circular region which is
hounded by two radii, EA, FB, and an arc AB of the circle. We call H
a sector of the circle. A more precise definition follows.
Hgurc 1423
650 Circumferences and Areas of Circles
Chapter 14
Definition 14.12 (See Figure 1424.) Given a circle of ra
dius r with center P and an arc AB of this circle, the union of
all segments T<$ such that Q is a point on arc AB is called a
sector. We call AB the arc of the sector and we call r the
radius of the sector.
Figure 1424
Suppose that we arc given u sector of a circle with radius r and
center V as shown in Figure 1425. Let AB 1*? the arc of the sector.
For n > 2, let P h i*2,  • . P»_i be n  1 points on SB such that each
of the n angles
lAVP lt £P 1 VP*... 1 £P m _ l VB
has a measure of — • mAB, (In Figure 1425, n = 4.)
Figure 1425
A
""^T^J"
i y\
r
\p*
14.5 Are** of Cbcitft; Arc Ungth; Sector of a Orcle 651
Let
(1) An = APi + Fa?* + ■ • + PniB.
Let S., be the area of the polygonal region VAP1P2 • •*!* l®l th en
(2) Sn = 4fl a (AJ>i) + J*p!A) + • ; ■ 4 i««(P n iB),
where a n is the altitude to each of the bases 3JV F^T\ P*iR ©f
triangles AAVPi, AP1VP2, • • • , AP n _iV7J, respectively. From Equa
tion (2) we get
(3) % = ^niAPi + PjPa + ■ .. + P„iR);
hence
(4) S H = fynA*.
by substitution from Equation (1) into Equation (3).
We can make the difference between S„ and S (the urea of the
sector) arbitrarily small by choosing n large enough. Similarly, the dif
ference between a n and r can be made arbitrarily small by choosing n
large enough. Tt seems reasonable, then, that
Urn S n = S and lim o„ = r.
By Definition 14.10, lim A n = L, the length of arc Afr By applying
parts 3 and 4 of Theorem 14.12 to Equation {4) t we obtain
(5) lim S„ = Jiim 0*3,,
From equation (5), it follows that S = \tL is a formula for die area of
a sector with radius r and arc length h, We state this result as our next
theorem.
THEOREM 14. J 7 The area S of a sector is onehalf the product
of its radius t and the length L of its arc, that is,
S = «L
If we combine the results of Theorems 14.16 and 14.17, we get the
following theorem.
THEOREM 14,18 If M is the degree measure of the arc of a sector
with radius r, then the area «S of the sector is U==s } aw®* that is,
\,jou/
Proof: Assigned as an exercise.
€52 Circumferences and Areas of Circlet Chapter 14
Example 3 Find the area of a sector with radius 10 if the degree meas
ure of the arc of die sector is 72,
Solution: If we use the formula of Theorem 14.1 8 with r = 10 and
M = 72, we obtain
Note that, in the formula of Theorem 14.17, if L is the circumfer
ence C of the circle, then
L = %<nr
in id
S = ^r • 2frr = wr 2 ,
as it should for a circle, Similarly, in the formula of Theorem 14.18, if
the given arc is the circumference of the circle, then M = 360 and
S = (Jlftn* = ««,
as it should for a circle.
EXERCISES 14.5
■ In working title exercises of this set, do not use an approximation for w un
less instructed to do SO.
L Prove Corollary 14.14.1. Let Sj and Sv be the areas of two circles with
radii r t and r%, respectively. Prove that
A. Si) ? fe*, *&
2. The area of one circle is \ times the area of a second circle. What is the
ratio of the radius of the first circle to the radius of the second circle?
3. The radii of two circles are r and 2r, How does the area of the larger
circle compare with the urea of the smaller circle? How does the cir
cumference of the larger circle compare with the circumference of the
smaller circle?
4. Show that if r < 2, the area of a circle is less than the circumference of
the circle. (Of course, the urea units and the length units are in different
systems. If the length units are inches and the area units arc square
inches, and if r < 2, then the statement to be proved asserts that the
number of square inches in the area is less than the number of inches in
the circumference.)
14,5 Aran of Circl**: Arc Ltngth; Stctor of a Clrcla 653
In Exercises 511, r is the radius of a circle and S is its ana.
5. If r = 8, find & 9. If r = y/4\, find S.
6. If S = \mv, find r 10. If r = x/TTir. find S,
7. If S = 154, find r. U. If S = 75.30, find r.
8. If r = 3ir, find S.
12. In Exercise 7, use ir  If and find r,
13. In Excrdsc 9, use ?r = 3.14 and find S to the nearest tenth.
14. hi Exercise 11. use w = 3.14 and find r to tlie nearest tenth.
In Exercises 1518, the area of a drab is given. In each exercise, find die
circumference of the given circle,
15. 81*r 17. 81
16. 49w 18. 49
In Exercises 1922, the circumference of a circle Is gfven. In each exercise,
find the area of the given circle.
19. 40*r 21. 40
20. lfor 22. 16
23. The figure shows two concentric circles with radii 7 cm, and 10 cm. The
union of the two circles and the shaded portion between them is some
times called an annul us. Find the area of the annulus.
24. A square and a circle have the same area.
(a) Find the perimeter of tlie square in terms of the radius r of the circle
(b) Find the circumference of the circle In term* of the length i of the
side of the square.
25. Find the ureas of the inscribed and curcuinscribed circles of a square
with side length 12.
26. In Exercise 25, find die area of the annulus {see Exercise 23) between
the two circles.
654 Circumferences and Areas of Circle* Chapter 14
27. The radius of the circle with center C is 24. The radius of the circle
with center C is 12, If mX3=60 and mA'fi'=60 T show lAB^ZlAll'.
.A
28. For the circles of Exercise 27, if IAB = lA'B' and mAB = 60, find
mA'B'.
Given a circle with radius r and an arc AB of the circle, use the given infor
mation in Exercises 2935 to find the indicated measure.
29. If r = 12 and mAB = 45. find IAB,
30. If r = 1 and IAB = ^ t find mAB.
4
31. If /AB = 11^ and mAJI = 60, find r.
6
32. If LAB = ^L and mA3 = 270, find r.
33. If r = 18 and mAB = 240, find /AB.
34. Kr=9 and mAB = 132, find IAB,
35. If r = 4 and fAH = Git, find iriAB.
36. Given a circle witli radius fa 1, find the degree measure of each of the
following arcs of the circle.
a) AB tf IAB = £
' 6
(b) .4C if IA 1; = £
4
(c) ADiflAD = Z
(d) AE if iH = J
(e) AF if 1AF = £
3
t AC if 1AC  $f
4
(g) ah if zaT/ = ^
6
(h) A/if 7A/ = ff
(i) A/ifJAJ = ~
o
;j) AK if lAK = ^
(k) Af,ifZAL =
4c
AM if 1AM =
(m) AJV if IAN = 2f
(n) /IP if LA? = ^
(o) AQ if &® = ll£
14,5 Areas of Clrct««i Arc Length; Sector of a Circle 65S
37. Prove Theorem 14,18, (Hint: Combine the results of Theorems J 4, 16
and 14.17.)
38. The radius of a circle is 12 in, Find the area of a sector with the follow
ing arc length:
(ft) 4r; (b) 2.7* (c) 8w (d) ?
39. The radius of a circle is 15 in. Find the area of a sector with an arc whose
degree measure is
(a) 60 (b) 144 (c) 1 (d) 330
40. Let a circle with center P and radius r be given. Points A and B arc
points of the circle such thai m Z A PB = 120 and the area of sector A PB
is 12ir. Find r and /AB.
41* In an xyplaue, let sets C and I be defined us follows:
C = {(.t, y) : x* + t/ a ■ 64},
I = {{x, u):ij = x).
Let P be the point where / intersects C in the first quadrant, let O be the
origin of the given xyplane, and let B he the point cm the positive ucaxis
where C intersects the xaxis. Find the area of the sector POB.
Exercises 4244 refer to Figure 1426 which shows a circle with center V*
and chord AB. The shaded portion hounded by the chord AB and the arc
AB is called a segment of the circle. l.et h he llw altitude to AB of A AVB,
let AB = s t let VA = r = VB, and let L  1A&.
Figure 1426
42. challenge problem. Derive a formula for h in terms of r and .?.
43. challenge problem. Derive a formula for the area S of the segment
of the circle in terms of r, $ t and L
44. See Exercise 43. Find S if r = 5, s = fl, and L = ~. (Use n = 3.14
and round your answer to the nearest tenth.)
656 Circumferences and Areas of Circlet
Chapter 14
45, challenge problkm. The figure shows three congruent circles with
centers P, Q, R. Each of the circles is tangent to the other two and the
points of tangency are A, B„ C,
(a) Show llutt AP{)R is equilateral.
(b) If the radius of each of the circles is 12, find the area S of the shaded
region bounded by arcs AB t BC, and AC. (Use it = 3.14 and round
your answer to the nearest tenth.)
CHAPTER SUMMARY
In this chapter we defined the following terms and phrases. Be sure that
you know the meaning of each of them.
INTERIOR ANGLE OF A
POLYGON
EXTERIOR ANGLE OF A
POLYGON
REGULAR POLYGON
( JrU I \IS( H IRED CIRCLE
(CIRCUMCLRCLE)
INSCRIBED POLYGON
INSCRIBED CIRCLE
(INCIRCLE)
CIRCUMSCRIBED POLYGON
CENTER OF REGULAR
POLYGON
CIRCU M RADIUS OF
REGULAR POLYGON
INRAD1US OF REGULAR
POLYGON
CENTRAL ANCLE OF
REGULAH POLYGON
CENTRAL TRIANGLE OF
REGULAR POLYGON
CIRCUMFERENCE OF
CIRCLE
THE NUMBER v
CIRCULAR REGION
AREA OF CIRCLE
LENGTHS OF ARC
SECTOR
ARC OF A SECTOR
RADIUS OF A SECTOR
Chapttr Summary 657
There were 18 theorems in tliis chapter, most of which consisted of
formulas. In developing many of these formulas, we used the idea of a
LIMIT of a sequence of numbers. A complete treatment of limits is too
difficult for a first course in geometry. Our goal was to give an intuitive
feeling for the concept <jf a limit and to make the formulas seem plausible,
Be sure that you know the following list of formulas and that you know how
to apply them.
Sum S of the measures oftiie interior angles of a convex polygon ofn
sides:
S s (n  2)180,
Measure m of each angle of a regular polygon of n sides:
, , ft  2 > 18Q ,
n
Sum S of the measures of the exterior angles, one at eacli vertex, of a
convex polygon of n sides:
S = 360,
Measure m of each exterior angle of a regular polygon of n sides:
^360
ft
Perimeter of a regular polygon of n sides:
p = 11%
where p is tlw perimeter and s is the length of a side of the polygon.
Area of a regular polygon of n sides:
S = iap,
whew S is Hie area, a is the inradhis, and p is the perimeter of the polygon.
Circumference of a circle:
C = vdorC = 2m,
where V is the circumference, r is tlte radius, and d it the diameter of the
circle.
Area of a circle:
where S is the area and r is the radius of the circle.
Length of an arc of a circle:
where L is the length of tlw arc, M i* the degree measure of the arc, and r
is the radius of the circle.
Area of a sector:
S = ±rL,
where Sis the area, r is the radius, and L is the length of die arc of the sector.
ess Circumferences and Area* of Clrcln Chapter 14
REVIEW EXERCISES
1. Find the measure of each interior angle of a regular pentagon; of a
regular 7gon.
2. Find the measure of each exterior angle of a regular 15gon.
3. If the measure of each interior angle of a regular ngon fa 150, find n.
4. If the measure of each exterior angle of a regular ngon ii 40 t find n.
5* If the measure of each of the n central angles of a regular ngon is 20,
find n.
6 Using only a pair of compasses for drawing circles and a straightedge
for drawing lines, explain how yon would construct a regular hexagon;
an equilateral triangle? a square.
7. Find the perimeter and the area of a regular hexagon each of whose
sides is 20 cm, long.
In Kxercises 813, the radius of a circle is given. In each exercise, find the
area and the circumference of the circle. Use tt = 3.14 and express each
answer to the nearest tenth,
8. r = 13.5 cm.
9, r= 1
10, r = 6.04 in.
11. r= 3,14 ft,
12. r = 12
13, r = 62.8
In Exercises 1419, the circumference Oof a circle or the area S of a circle
is given. In each exercise, find the radius of the circle. Give an exael answer
in each case.
14. S = 64w
15. C = 64*
16. S = 216V sq. cm.
17. S = 225 sq. in.
18. C = 14&7 ft
19. C = 72 cm.
In Exercises £024, r is the radius of a circle, L is the length of an arc of the
circle, and M is the degree uieusure of the arc.
20. If r = 16 and M = 80, find L.
21, If r = 4 and L = 3ir t find hi
Rtvlvw Extrclttt
659
22. If L = ^ and M = 240. find r.
23. If r = 10 and L = i^, find M.
24. ff r a 45 and M m 220, find L.
25. An annulus (the shaded region shown in the figure) has an inner radius
of x and an outer radius of x + 3, If its area is 48?r. find its inner and
outer radii.
Iii the figure, AC?, BC. fixU, and ffiB are semicircles, with A, U. C
collinear. If AC = BC = 7, find the area of the shaded region.
fa..?.)
£7. Find the area of a sector if the radius is 14 and if the length of the arc
of the sector is lit;.
660 Circumferences and Areas of Circles Chapter 14
28, In the figure, /\ABC is ft right triangle with the right angle at C. An,
BV, and AC arc semicircles wilh diameters AB t BC : and AC", respec
tively. If a, b, c arc the lengths of the sides EC, KTL AB, respectively,
show that the area of the semicircle with diameter c is equal to the sum
of the areas of the two semicircles with diameters a and b*
29. In the figure, A ABC is a right triangle with the right angje at C, KB is
a diameter of circle JfC, and AKC and BYC arc semicircles with diam
eters AC and BC. respectively. Show that the sum of the areas of the
two shaded regions is equal to the area of the triangle
Review Exercises 661
30. CBHJUJKN&B raoBX.EM. In the figure, A BCD is a square euch of whose
sides is 10 cm. long, P, Q t R t S are the midpoints of sides A~R, fi€, CD,
DA, respectively. Arcs EF, FC, Oil, HE are arcs of circles with centers
P, Q, R, S, respectively, and are tangent to the diagonals of square
ABCD at points E, P, G, If.
(a) Prove that PQRS is a square.
(b) Find die area of the shaded region bounded by the four arcs EF,
FG t Gtf , and HE.
Vim Bucher/Fhoto Researchers
Areas
and Volumes
of Solids
15.1 INTRODUCTION
In your Study of informal geometry you may have learned formulas
for finding surface areas and volumes of some of the familiar solids.
In this chapter we review and extend this phase of geometry We con
tinue our somewhat informal development of geometry from Chapters
13 and 14. A formal development of these formulas belongs properly in
a calculus course. The development of this chapter is designed to make
the formulas plausible. Emphasis is placed on understanding the for
mulas and on using them.
In Chapter 13 we studied circles and spheres, A sphere may be
thought of as a solid or as a surface. In this book a sphere is a surface,
and tlie union of a sphere and its interior is a spherical region. In
formally speaking, the area of a sphere is a number that expresses the
measure of the sphere, and the volume of a sphere is a number that ex
presses the measure of the associated spherical region.
Although prisms, pyramids, cylinders, and cones may be thought of
as either solids or surfaces, in this book we consider them as solids. The
664
Areas and Volumes of Solids
Chapter 15
area of a cylinder (more properly., the surface area of a cylinder), for
example, is a number that expresses the measure of the surface of the
cylinder, and the volume of a cylinder is a number that expresses the
measure of the cylinder itself,
15.2 PRISMS
Figure 151 shows some diagrams of prisms. We might think of a
prism as the solid swept out by a polygonal region moving parallel to
itself from one position to anodier. Each point P in the region moves
along a segment PF as suggested in Figure 152, and all of these seg
ments are parallel to each other. The prism is the union of all such seg
ments. Wc make these ideas formal with the following definition.
Figure 151
15.2 Prisms
.■53
Definition 15.1 ;'See Figure 152.) Let a and ft be distinct
parallel planes. Let Q and Q* be points in a and ft, respec
tively. Let R be a polygonal region in a. For each point P in
R let F be the point in ft such that PF  QQ*. The union of
all such segments PF is a prism, if QQ' is perpendicular to
a and ft, die prism is a right prism.
Figure 154
In this chapter we shall limit our discussion of prisms to those
prisms whose bases are convex polygonal regions, that is f regions whose
boundaries are convex polygons*
Definition 15,2 (See Figure 152.) Let R r be the polygonal
region consisting of all the points F in ft. The polygonal re
gions R and R' arc called the bases of the prism, Depending
upon the orientation of the prism it is sometimes convenient
to call one of the bases the lower base and the other base the
upper base. Sometimes we call the lower base simply the
base. A segment that is perpendicular to both a and ft and
with its endpoints in these planes is an altitude of the
prism. Sometimes the length of an altitude is called the alti
tude of the prism.
Prisms are often classified according to their bases. Thus a triangu
lar prism is one whose base is a triangular region; a rectangular prism is
one whose base is a rectangular region, and so on.
65ft
Ar#av and Volumes of Solids
Chapter 15
Figure 153 shows a triangular prism. The bases are the triangular
regions ABC and A'B'C. The triangular region DEF in the figure,
which lies in a plane parallel to the plane of the base, is called a crass
section of the prism.
\
Figure 153
Definition 15.3 If a plane parallel to the plane of the base
of a prism intersects the prism, the intersection is en' led a
cross section of the prism.
Is the triangular region ABC a cross section of the prism shown in
Figure 153? Is the region A'B'C a cross section of the prism?
We say that a point P of the lower base corresponds to a point R of
a cross section (other than the lower base) if PR , Qty. We call A DEF
in Figure 153 the boundary of the triangular region DEF. Similarly.
A ABC is the boundary of the lower base of the prism, & A'B'C is the
boundary of the upper base, and so on.
THEOREM I5>] The boundary of each cross section of a tri
angular prism is congruent to the boundary of the base of die prism.
Proof: Let the triangular region ABC in plane a be the lower base of
the prism and the triangular region A'B'C in plane j8 be the upper base
as shown in Figure 153. Let y be a plane parallel to a and intersecting
AA', BB' t CO in points D % E, F, respectively. Then, by definition, the
region DEF is a cross section of the prism. That the region DEF is a
15.2 Prism* 667
triangular region can be proved using separation properties. We omit
the details here. We shall prove ADEF « A ABC.
If y = a, then D = A, E ! = H. F = C, and ADEF = AABC.
Therefore ADEF s AAiiC. Why? Suppose, then, that y=£aas sug
gested In Figure !53. Bv the definition of a prism, there are points Q
in a and Q'Jn p such that A A' \ QQ' and BB' \ QQ'. Then AA'  BB'
and AD  BE. Also, by Theorem 8.12, Aft I DE. Therefore ABED is
a parallelogram and 0E == AB. In the same way we can prove
W z. AC and FE Sk CB. Therefore A DEF s AABC by the S. S. S.
Postulate. Since y is an arbitrary plane parallel to a and intersecting
the prism, we have proved that the boundary of each cross section of a
triangular prism is congruent to the boundary of its base, and the proof
is complete.
Since the upper base of a prism is a cross section of the prism, we
have the following corollary.
COROLLARY 15.1.1 The boundaries of the upper and lower
bases of a triangular prism are congruent
THEOREM 15.2 (The Prism Crow Section Theorem) /Ml cross
sections of a prism have the same area,
Proof: IM the prism as shown in Figure 154 be given. (We have
shown a prism whose base is a polygonal region consisting of five sides.
The following argument can be modified to apply to a prism whose
base is a polygonal region consisting of n sides, where n > 3.) Let It be
the base and let R' be a cross section of the
prism. Then R can be divided into nonover
lapping triangular regions 6), £2, *3 as shown
in Figure 154, Let t' lt & g be the corre
sponding triangular regions in B\ Then the
boundary of h is congruent to the boundary
of fj, the boundary of h is congruent to the
boundary of t 2 , and the boundary of 1 3 is con
gruent to the boundary of £3, Why? The areas
of t[, &, t'3 are equal, respectively, to the areas
0* h, h, k Why? The area of R f is the sum
of the areas of r[, if, tk, and the area of R is
the sum of the areas of tu tz, h Why? Since these two sums are equal,
it follows that the area of R' is the same as the area of R. Since R' is an
arbitrary cross section of the prism, it follows that all cross sections
have the same area and die proof is complete.
668 Areas and Volumes of Solids Chapter IS
COHOLLARY 15.2.1, The two bases of a prism have die same
an. a,
Proof: Assigned as an exercise.
Definition 15.4 (See Figure 155.) A lateral edge of a prism
is a segment AA'. where A is a vertex of the base and A' Is the
corresponding vertex of the upper base. Given any side of
one base of a prism, the lateral face of the prism correspond
ing to that side is the union of all segments FF parallel to a
lateral edge and with P on the given side of die base. The
lateral surface of a prism is die union of its lateral faces. The
total surface of a prism is the union of its lateral surface and
its bases.
Hgu» 155
For tlie prism shown in Figure 155, AA' is a lateral edge and
AHB'A' is a lateral face. Name four other lateral edges and four other
lateral faces in the figure.
Note that die lateral edges of a prism are parallel to the segment
QQ' in the definition of a prism. The segment AB in Figure 155 is not
a lateral edge of the prism although AB is an edge of the prism. Indeed,
it is an edge, or side, of a base of the prism.
THEOREM 15,3 The lateral faces of a prism are parallelogram
regions and the lateral faces of a right prism are rectangular regions.
Fmvfi (See Figure 155.) A complete proof involves a discussion of
separation properties which we omit here. Suppose that we are given a
15.2 Prisms 669
prism with points labeled as in Figure ,155. We shall content ourselves
with proving that ABB'A' is a parallelogram and that ABB' A' is a rec
tangle if the prism is a right prism. Similar arguments could he given for
each of the lateral faces of the prism.
By the definition of a prism, there is a segment (5p' such that
AA'  QQ' and M' QQ' Therefore Id' \\ BB'. Why? By the defini
tion of a prism, the planes containing the bases are parallel. Therefore
it follows from Theorem 8.12 that AH'  AB. This proves that i\BB'A'
is a parallelogram.
If die prism is a right prism, then ^?' is perpendicular to plane a.
Why? Therefore AA'' a (Why?) and AA"' I AB. Why? It follows
that ABB' A' is a rectangle.
Definition 15.5 A parallelepiped is a prism whose base is
a parallelogram region. A rectangular parallelepiped is a right
prism whose base is a rectangular region. A cube is a rec
tangular parallelepiped all of whose edges are eongruenL A
diagonal of a parallelepiped is a segment joining any two of
its vertices which are not contained in the same lateral face
or tiase of the parallelepiped.
Figure 156 shows pictures of a parallelepiped, a rectangular par
allelepiped, and a cube* In each picture, the segment rlS is a diagonal
of the parallelepiped. How many diagonals does a parallelepiped have?
.'.,!„
Rrc'juifiuUr
Parallelepiped purillatfpspftd
an"
*■
'
X
1 II
_J \
Figure lofi
Definition 15.6 The lateral surface area of a prism is the
sum of the areas of its lateral faces. The total surface area of
a prism is the sum of die lateral surface area and the areas of
die two bases.
670 Areas and Volumes of Solids
Chapter 15
EXERCISES 15.2
1. Copy and complete;
All the faces of a parallelepiped (lateral,
upper base, and lower base) are [?]
regions.
2. Copy and complete:
Alt the faces of a [JJ parallelepiped are
rectangular regions,
3. Copy and complete:
All the faces of a [JJ are square regions.
Exercises 416 refer to the prism shown in Figure 157. In this figure. Pis a
point in a and V is a point in /? such thai. PF JL a. In each exercise, com
plete the statement.
4. The region ABCDE is called a [JJ of die prism.
5. The region A'lfCtXE' h caBod u [JJ of the prism.
G. BB' is called a (an) JJ of the prism.
7. There are [Jj lateral edges in all.
8. Counting the lateral edges and the edges (sides) of the two bases t there
are [jj edges in all.
9. The parallelogram region BCCB' is called a (art) [JJ of the prism,
10. There are [JJ lateral faces in all.
15,2 Print* 671
It, Counting the lateral faces and the two liases, there arc [T feces in all
12. A is called a (an) [T) of the prism.
13. There are (TJ vertices in all
14. If V is the number of vertices, £ is the numher of edges, and F is the
number of faces, then V — £ j F = [T],
15. PP is called an T of the prism.
16. If M' _L 0. then the prism is called a {?}.
17. Prove that the total surface area of a cube is 6e 2 , where e is the length
of one of its edges.
IS. Fmd the total surface area of a cube whose edge is 7 cm. in length.
19. Prove that the total surface area of a rectangular parallelepiped is
lab + 2lic + 2uc,
where a and h are the dimensions of the base and c is the altitude of the
prism. Draw an appropriate figure,
20. Find the total surface area of a rectangular parallelepiped if the dimen
sions of the base are 4 cm. by 6" an. and if the altitude of the prism is
8 cm,
21. Given that the pentagonal prism of Figure 157 is a right prism, that the
lengths of the edges of the base are 3, 7, 4, 9 1 and 6, and that the alti
tude is S, find the lateral surface area of die prism.
22. If S is the lateral surface area, a is the altitude, and p is the perimeter of
the base of a right prism, prove that
S = ap.
23. Use the formula in Exercise 22 to find the lateral surface area of tiia
prism in Exercise 21. Does your answer agree with the one obtained in
Exercise 2 1 ?
24. Find the altitude of a right prism if the lateral surface area is 336 and
the perimeter of the base is 28.
25. Find the perimeter of the base of a right prism if the lateral surface area
is 351 and the altitude is 13^.
26. If the base of the right prism in Fxercise 25 is a square region, find the
length of each of its edges.
27* Find the total surface area of a right triangular prism if the boundary
of each base is an equilateral triangle whose sides have length 10 and If
the altitude of the prism is 12.
2S. Ilie area of a cross section of a prism is 32. The lateral surface area is
128 Find the total surface area of the prism.
29. If XB and FQ are two lateral edges of a prism, prove that AB and TQ
arc eoplanar.
672 Areas »r»d Volumes of Solids
Ctiapler 15
30. Prove Corollary 15.2 J,
31 , Given the rectangular parol lelepiped shown in the figure with AB = 12,
BC = 6, and CD = 8, find the length of die diagonal Ail, (Hint: Draw
At?. What kind of triangle is A ABC? What kind of triangle is AACD?)
32. In the figure RS is a diagonal of a cube and the length of each edge of
the cube is 8. Prove that RS = 8^/5".
33. Prove that the length of every diagonal of a tube is e y/S, where e is the
length of one of its edges.
34. challenge problem. Use the Distance Formula for a threedimen
sional coordinate system to prove that the diagonals of a rectangular
parallelepiped have equal Lengths.
35. challenge PROBLEM. If h is the altitude of a prism, prove that h <, r,
where r is the length of any one of its lateral edges.
15.3 PYRAMIDS
Figure 158 on page 673 shows some pictures of pyramids. Com
pare Figure 158 with Figure 151, In what respect does a pyramid
differ from a prism? How arc they similar?
Since a pyramid is similar in many respects to a prism , some terms
for parts or' a prism are also used for parts of a pyramid. We shall use
these terms without giving formal definitions.
15.3 Pyramids 673
Figure I5^i
Definition 15.7 (See Figure 159,) Let Rka polygonal re
gion in a plane o and V a point not in a* For each point P of
R there is a segment FV. The union of all such segments is
called a pyramid. The polygonal region R is called the hase
and V is called the vertex of the pyramid. The distance VT
from V to a is the altitude of the pyramid.
tx
Figure 15*9
For the pyramid in Figure 159, AV is a lateral edge and the tri
angular region ABV is a lateral face, Name four other lateral edges
and four other lateral faces of the pyramid shown. How many edges
lateral and base) does Ida pyramid ki\v m ,,11 IIow main laces in all?
How many vertices in all?
A cross section of a pyramid is the intersection of the pyramid with
a plane parallel to the l*ase provided the intersection contains more
than one point.
674 Areas and Volumes of Solids
Cfneiir 15
THEOREM 15.4 The boundary of each cross s cction of a tria agu 
tar pyramid is a triangle similar to the boundary of the base, and the
areas of any two cross sections arc proportional to Lhe squares of
the distances of their planes from the vertex of the pyramid.
Proof: Tjet the triangular region ABC in plane a he the base of the
pyramid as shown in Figure 1510. LcL fi be a plane parallel to a and
intersecting AV, BV, and EV in distinct points A\ W t and C. respec
tively. Then the triangular region A'B'C is a ctoss section gf the pyra
mid. Let S be the area of A ABC, let S' be the area of AA'B'C, let Jt be
the distance from the vertex lo the cross section plane, and let h be the
altitude of the prism. In Figure 1510, Jt = VF and k = VP.
k = VP'
h = VP
Figure 1510
To complete the proof of the theorem we shall prove statements
1 and 2.
L AA'B'C ~ AABC
2. (S\S) = (**,**)
If/? = «,lhenA'=A.H' = B, C = C, Y = F t and k = h. There
fore AA'B'C  AABC and hence
A A'B'C s AABC.
It follows that AA'B'C'  AABC and that
S' = S, k = h, and (S', S) = (&*, *■).
Suppose, then, that £ ^ «. We have V 2 A, F A VF coplanar (Why?)
with AA'V and FFV. AT ± Wand A*F i VF. Why? Therefore
15.3 Pyramid* 675
AA'FV A APV by the A.A. Similarity Theorem. (lA'VF^lAVP
LA'VV= ZAPV/i Hence
(VA' t VA) = {k. h).
In the same way, we can show that AB'FV — ABFV and hence
(VB, VB> = (*, /i).
Therefore
£VA' T VA) = (VB' t VB)
and
AA'W  AAVi*
by the S. A, S. Similarity Theorem. Therefore
(A'B',AB) = (VA\VA)
and
I A'B\ AS) = (ft, h)
A'B' = 4'AJ3.
In the same way it can be shown that
ft
h
= £*8C
and
CA' =
Then
C'A' = ^>CA.
{A'B f , B'C, CA') = (AB, BC, CA)
and it follows that AA'B'C < AABC by the S. S. S. Similarity Theo
rem. Recall that in Chapter 10 it was proved (Theorem 10.15) that if
two triangles arc similar, then their areas are proportional to the square
of the lengths of any two corresponding sides. Therefore
(S\ S) s ((A'/?)*, (AB)*).
Since
(A'B 1 , AB) = (ft, h) t
it follows that
{(A'l?)\ (AB)*) = (ft*, V).
Therefore
(S\ S) = (ft». M)»
and the proof is complete.
676 ATM* and Volumes of Solids
Chapter 15
Compare our next theorem for pyramids with Theorem 15.2 for
prisms.
THEOREM 15.5 In any pyramid the areas of any two cross sec
lions arc proportional to the squares of the distances of their planes
from die vertex of the pyramid.
Proof: Let a pyramid be given as shown in Figure 151 1, (We have
shown a pyramid whose base is a polygonal region consisting of five
sides. The following argument can be modified to apply to a pyramid
whose base is a polygonal region consisting of n sides, where n > 3.)
k m VP'
h=VP
Figure 1511 B
Let the polygonal region ABCDE be the base and let A'WC&W
be any cross section of the pyramid* Let S be the area of Ihe region
ABCDE and let S' be the area of the region A'B'CD'E'. Let k be the
distance from the vertex to the cross section plane and let h be die alti
tude of the pyramid. Hie region ABCDE can be divided into nonover
lapping triangular regions tu t 2 , h as shown in Figure 151 L Let t{, t' 2 ,
t$ be die corresponding triangular regions mA'E'CD'F/. Let Tt. 7 a, T s
be the areas of t u r 2 , # y , respectively, and let If, Ti, 'ft be the areas of
t{, <2» fii, respectively.
Thci
s = n + n + r 3
S = T, + T 2 + T;
3
15.3 Pyramids 677
By Theorem 15.4,
From the product property of a proportion it follows that
Tift* = TiR
Similarly,
and
T 2 h* = T 2 fc*
r s h* = raft?.
Then adding and using the Distributive Property, we get
pi + n + n)h £ = (Tt + r 3 + rajtf
S'Ji* = Sfc*
(S\ S) = (fc 2 , **)
and this completes the proof.
We now use Theorem 15.5 to prove our next theorem.
THEOREM 15.6 (The Pyramid Crow Section Theorem) If two
pyramids have equal altitudes and if their bases have equal areas,
then cross sections equidistant from the vertices have equal areas.
Proof: Let two pyramids be given as shown in Figure 1512. (We have
shown pyramids whose bases are triangular regions. The theorem, how
ever, is not restricted to this case and our proof applies equally as well
to pyramids whose bases are polygonal regions with more than 3 sides.)
Figure 1!S 12
Areas and Volumes of Solids
Chapter 15
Let the base area of each pyramid be S, let h be the altitude of each, and
let k be the distance from the vertex to the plane of the cross section of
each. Let the areas of the cross sections be Si and S2. We must prove
that Sj = Sg. By Theorem 15.5,
Therefore
(Si. S) f (*t, A*) = (fig, S).
Si = s 2 ,
and the proof is complete.
EXERCISES 15.3
In Exercises 113, copy and complete each statement. Exercises 313 refer
Lo Ore pyramid shown in Figure 1513 in which P is a point in a, the phi n ltd
the base, such that f la,
Figure 15 IS
1. All the lateral faces of & pyramid are [7] regions.
2. If the boundary of die base of a pyramid is an equilateral triangle, then
the boundary of each cross section is Tj.
3. The region ABCD is called the ]T] of the pyramid.
4. V is called the G] of the pyramid.
5 A V is called a [?) of the pyramid*
6, There are [?] lateral edges in all
7. Counting the lateral edges and the edges of the bases, there are \T\ edges
in all.
8. The triangular region AAV is called a of the pyramid.
9, There are [fj lateral faces in all.
15.3 Pyramids 679
10 Counting the base, the total number of faces is [TJ.
11. There are \J} vertices in all.
12. If Vis the number of vertices, K the number of edges, and F the number
of faces, then V  E + P = 7J.
13 VP is the [T] of the pyramid,
14. Compare your answer to Exercise 12 with that of Exercise 14 in Exer
cises 15.2. Are they the same?
15. Compute V  E + F(scc Exercise 12) for the pyramid shown in Figure
1511 and for the prism shown in Figure 155, Are your answers the
same for both solids? Do you think V — E + F = 2 for every prism or
pyramid? Try some more examples using figures from this chapter.
16. Recall that the center of a regular polygon is the center of the circum
scribed circle. The center of a regular polygonal region is the center of
the polygon hounding the region. A pyramid is a regular pyramid if and
only if its base is a regular polygonal region and the foot of the perpen
dicular from its vertex to its base is the center of the base. Figure 1513
shows a regular pyramid whose base is a square. Prove that the bound
ary of the lateral face A VB is an isosceles triangle. (Hint: Draw PA and
PBand prove that &AFV m ABPVby the S. A. S. Postulate.)
17. Given that the pyramid shown in Figure 1513 is a regular pyramid
whose base is a square region (see Exercise 16), prove that
AAVB m ABVC.
18. Draw a picture of a regular pentagonal pyramid, that is, a regular pyra
mid whose base is a regular pentagonal region (see Exercise 16). Label
the vertex Vand the vertices of the base A, B t C, D t E. Pick any two
lateral faces and prove that they arc congruent isosceles triangles. (Hint:
Draw the perpendicular from V to the center of the base as in Figure
1533.)
19. Recall that corresponding altitudes of congruent triangles are con
gruent. Prove that the lateral surface area of the regular pyramid shown
in Figure 1513 is given by S = £ap, where p is the perimeter of the
base and a is the length of the segment whose ondpoints are the vertex
of the pyramid and die fool of the perpendicular from the vertex to an
edge of the base, (See Exercises 16, 17, and 18.)
20. Let two pyramids, one triangular and one hexagonal, with equal base
areas be given. The altitude of each pyramid is 9 in. The cross section of
the triangular pyramid that is 3 in. from the base has an area of 40 sq. in.
What is the area or the cross section of the hexagonal pyramid that is
3 in. from its lrase?
21. The area of the base of a pentagonal pyramid is 1024. The distance from
the vertex to the plane of a cross section is 3 and the altitude of the pyra
mid is 8. Find the area of the cross section. ; J Jink Use Theorem 15.5.)
680 Art** and Volumes of Solids Chapter 15
22, The boundary of the base of a pyramid is an equilateral triangle, and the
boundary of each lateral face is an equilateral triangle with sides of
length 8. Find the total surface area of the pyramid.
23, The boundary of the base of a pyramid is a square whose sides are
10 em. in length, and the boundary of each lateral face is an equilateral
triangle. Find the total surface area of the pyramid.
24, Find the altitude of the pyramid of Exercise 23,
25, Find the altitude of the pyramid of Exercise 22.
26, The area of a cross section of a pyramid is 125. The area of the base is
405. The altitude of the pyramid is 9. Find the distance k from the vertex
to the plane of the cross section.
fr. Given (a, b) = (c, d) t prove that (a*, fe 2 ) = (c 2 , <#). If
(<j, b) = (c, d)
with constant of proportionality t> then what is the constant of pro
portionality for
{«*, £*) = ( C 2, tP)?
Refer tr> the proof of Theorem 15.4. In which sentence of this proof is
this property of a proportionality used?
15.4 AREAS AND VOLUMES OF PRISMS AND CYLINDERS
As stated in the inrrixhn.i inn i<> this chapter, we shall not tcat areas
and volumes of solids as rigorously as we did areas of polygonal regions.
The concepts of surface area and volume are natural extensions of the
area concept developed in Chapter 9. We accept without proof the
fact that each solid discussed in this chapter has a surface area and a
volume.
For volumes, we accept two postulates and use them to prove the
volume formulas for prisms, cylinders, pyramids, and cones. Later in
the chapter we develop a formula for the volume of a sphere.
For surface areas of cylinders, cones, and spheres, we develop the
formulas informally and then state them formally as theorems whose
proofs are beyond the scope of this book.
Our first postulate is similar to Postulate 28, Lhe Rectangle Area
Postulate of Chapter 9.
POSTULATE 31 (Rectangular Parallelepiped Volume Postu
late) The volume of a rectangular parallelepiped is the product of the
altitude and the area of the base.
15.4 Areas and Volumt* of Prisms and Cylinders 681
Figure 1514 is a picture of a rectangular parallelepiped. By defini
tion, its base is a rectangular region and each of its faces is a rectangular
region. Thus any one of its faces could be called the base, and the
length of any one of its edges that is perpendicular to that face could be
called the altitude.
Hfur* 1544
By Postulate 31, the volume Vof the rectangular parallelepiped shown
in Figure 1514 is given by the formula
V=Sh,
where S is the area of the base and h is the altitude. By Postulate 28,
S = oh. Therefore we have
V = abh
as a formula for the volume of a rectangular parallelepiped.
Suppose that an ordinary deck of playing cards is arranged so that
its lateral faces are vertical as suggested in Figure 1515a. It seems rea
sonable dial if the shape of the deck is changed as in Figure 1515b, the
volume of the deck remains the same. It also seems reasonable that the
fifteenth card from the bottom in Figure 15 15a has the same volume
as the corresponding card in Figure 1515b.
Figure IS IS
682 Areas and Volumes of Solids Chapter 15
As a second example,, suppose that we make an approximate model
of a rectangular pyramid by forming a slack of thin cards as suggested
in Figure 1516. Of course, the thinner we make the cards, the more
cards there will be and the closer the approximation will be. Suppose
that the cards in die model are kept at the same level but are allowed
to change position by sliding along each other.
Figure 1516
Then the shape of the model changes, but its volume does not change.
For different positions of the cards, the base area and the thickness
of any two cards at the same level are the same, hence their volume
is the same. The total volume of the model is the total volume of the
cards, and the total volume does not change when the cards slide along
each other.
More generally, imagine two solids with equal altitudes and with
bases in the same horizontal plane. Suppose that every two cross sec
tions of these solids at equal distances from the bases have equal areas.
Then it seems reasonable that the two solids should have equal vol
umes. The reason is that if we imagine die solids being eut into thin
slices by planes parallel to the bases, the volumes of slices at the same
distance from the bases will be approximately equal. Therefore the
volumes of the two solids should be equal.
The principle that we have tried to make plausible here is called
Gavalieri's Principle after Professor Bonaventum CavaKeri (1598
1647) of the University of Bologna. He used this principle m obtaining
some results that we now find in the calculus. We state this principle,
formally as the second postulate of this section.
POSTULATE 32 (Cflw/iWi Principle) If two solids have equal
altitudes, and if cross sections of these solids at equal distances from
the bases have equal areas, then the solids have equal volumes.
15.4 Area* and Volumes of Prisms and Cylinders 683
Cavalicri's Principle is the key to calculating volumes other than
rectangular parallelepipeds. We use the principle in the proof of our
next theorem,
THEOREM 15. 7 The volume V of any prism is the product of it?
altitude h and the area S of its base, that is, V = Sfc.
Let a prism with altitude h and base area S be given as shown
in Figure 1517a, Let the rectangular parallelepiped shown in Figure
15t7b have the same altitude h and the same base area S T and let its
base and the base of the given prism be in the same plane.
Figure 15*17
By the Prism Cross Section Theorem, all cross sections for both prisms
have the same area S. By CavalierTs Principle, the two prisms have the
same volume. It follows from the Rectangular Parallelepiped Volume
Postulate that V = Sh for the given prism.
Example I The base boundary of a prism is an equilateral triangle
8 in. on a side. Its altitude is 12 in. Find the volume of the prism.
Solution: The volume of the prism is given by V = $h t where S is the
area of the triangular base and h is the altitude. We have S = 16 \/3
(show this) and h = 12. Therefore
V = 16 V3*I2 = 192 V3,
and the volume is 192 v3"cu. in.
684
Areas and Volumes of Solids
Ch»pt«r 15
Figure 1518 shows a diagram of a circular cylinder. Many of the
terms used in defining a prism apply equally as well to a circular cylin
der. The base of a prism is a polygonal region.
Figure 151*
The base of a circular cylinder is a circular region, that is, the union
of a circle and its interior. Make appropriate changes in the wording
of the definition of a prism (Definition 15.1), refer to Figure 1518, and
define a circular cylinder.
There are other cylinders in addition to circular cylinders, that is,
cylinders whose bases are not circular regions; but we shall consider
only circular cylinders in this text. Therefore, when we speak of a
cylinder, we mean a circular cylinder.
If QQ' in Figure 1518 is perpendicular to a, then the cylinder is
called a right circular cylinder, The altitude, bases, and cross sections
of a cylinder are defined in (he same way as are the corresponding
parts of prisms.
The following two theorems are analogous to Theorems 15.1 and
15,2 for prisms and can be proved in a similar way. We omit the details
of the proofs.
T1IEOR FM 15.S Hi© bou udary of each cross section of a cyl i n der
is a circle that is congruent to the boundary of the base.
Outline of Proof: (See Figure 1519 on page 685.) Let R be the lyase
of the given cylinder and let H' be any cross section of the cylinder
Let C be the center of the circle that bounds the base, let P be a point
on that circle, and let C and F be the corresponding points in R' t Let
15.4 Areas «nd Volumes of Prisms and Cylinders 685
CP = t and let CF = t>, Then PCCF is a parallelogram (Why?), and
FC = r f = PC = r.
Since PC has a constant value regardless of the position of P on the
base circle, then FC has a constant value. Thus all points F lie on a
circle with radius r* and center U.
Figure 1519
Therefore the cross section is a circular region and its boundary is a
circle of radius r. Then its boundary is congruent to the boundary
of the base.
THEOREM 15.9 (The Cylinder Cross Section Theorem) The
area of a cross section of a cylinder is equal to the area of the base.
Proof: Assigned as an exercise*
Cavalieri's Principle is used in the proof of the following theorem on
the volume of a cylinder.
THEOREM 1 5 JO The volume of a cylinder is the product of the
altitude and the area of the Imss.
Proof: The proof is similar to that of Theorem 15.7 and is assigned as
an exercise.
Imagine slitting a right circular cylinder and unrolling its lateral
surface onto a plane. Which figure is obtained? Figure 1520 on page
081 BBggesis that the boundary of the region khufl obtained k a sec
tangle whoso altitude is the altitude of the cylinder and whose base
is the circumference of the base of the cylinder. Thus, if r is the radius
sac
Areas, and Volumes of Solids
Chapter 15
of the boundary of the base of a right cylinder and h is the altitude,
then the lateral surface area of the cylinder is equal to the area of the
rectangular region obtained by "unrolling" the cylinder, that is, the
lateral surface area is 2vrh, Since the area of each of the circular bases
is tff 2 , the total surface arm of a right cylinder is 27rrh + 2irA
L
Figure 1 53)0
We state these results formally as our next theorem.
THEOREM 15.11 The lateral surface area of a circular cylinder
of base radius r and altitude h is 2irrh and its total surface area is
2irrk + Swr 6 .
liliniiflb 2 Find the total surface area of a cylinder if the radius of
the circle that bounds the base is 7 and the altitude of the cylinder is 10.
Solution; The lateral surface area is
2irrh= 2ff7« 10 = Mto
The area of each base is
at* = ??(7) 2 = 49tt.
Therefore the total surface area is
140*7 + 249* = 140^ + 9&r = 23&r.
EXERCISES 15.4
1, A rectangular tank 6 ft, by 4 ft. is used for watering horses. If the tank is
filled with water to a depth of 3 ft., how many cubic feet of water are
in the tank?
2* One gallon of water occupies 231 cu. in, of space To the nearest hun
dredth, how many gallons of walcr are contained in a space of 1 cu. ft.'?
3. To the nearest gallon, how inany gallons of water are in the tank of
Exercise 1? (Sec Exercise 2.)
4, Show that the volume V of a cube of side e is given by V = *3.
15,4 Ar»a* and VoJumw of Prhmt and Cylinders 687
5. Find the volume of a cube whose edge is 6 in, Find its total surface area.
6. Write a formula for the volume Vof a cylinder if its altitude is h and the
radius of tts base circle is r.
7. A water tank in the shape of a cylinder is 40 ft. high. The diameter of its
base is 2S ft. Find the volume of the tank.
8. To the nearest thousand gallons, how many gallons of water will the
tank of Exercise 7 hold? (Use w = ^ and see Exercise 2.)
9. The altitude of a cylinder is 8 in. and the diameter of its ba.se circle is
3 in. Kind the volume and total surface area of the cylinder.
10. On a shelf in Roy's supermarket there are two cylindrical cans of coffee.
The first is 1 j times as tall as the second, but the second has a diameter
1^ times that of the first. How should Roy price the second can in rela
tion to the first if he wants the price per unit of volume to be the same
for both cans?
11. How do the volumes of two cylinders compare if their altitudes are the
same but the radius of the base circle of the second cylinder is three
times that of the first?
12. How do die volumes of two cylinders compare if the radii of their bass
circles are the same but the altitude of the second cylinder is throe
times that of the first?
13. Draw a suitable figure and prove Theorem 15.9.
14 I3raw a suitable figure and prove Theorem 15.10,
15. A brick chimney In the form of a cylindrical shell and 2o ft. tall is to be
built. The inside and outside diameters are 24 in. and 10 in., respec
tively. If it takes 31 bricks per cubic foot of chimney, find die approx
imate number of bricks needed (Use v = 3.14.)
16. An air conditioning unit is to be installed in a rectangular building. In
order to install the correct size unit, it is necessary to know the number
of cubic feet of air inside the building. If the dimensions of the building
are as shown in Figure 1521 , find the volume of air inside the building.
17. A block of wood in the shape of a cube has edges 16 in. in length. A
circular hole 7 in. in diameter is bored through the block from top to
bottom. Find die volume of the part of the block that remains.
18. In Exercise 17, if e is the length of die edge of the cube and r is the
radius of the circular hole, write ft formula for the volume Vof die block
that remains after the hole has been bored.
HI
Areas and Volumes of Solids
Chapter 15
15.5 VOLUMES OF PYRAMIDS AND CONES
Here wc develop formulas for calculating volumes of pyramids and
cones and for the surface area of a cone. As in Section 15,4 , Cavalieri's
Principle plays a key role in the proofs of theorems on volume.
THEOREM 15,12 Two pyramids with the same altitude and the
same base area have the same volume.
Proof: Let two pyramids lws given as suggested in Figure 1522. By
the Pyramid Cross Section Theorem, corresponding cross sections of
die two pyramids have the same area. Therefore* by Cavalier! s Prin
ciple, their volumes are the same.
Figure 1522
Our next theorem provides a formula for calculating the volume of
a triangular pyramid. Suppose that we are given a triangular pyramid
with base ABC and vertex W. (See Figure 1523a.) Next, wc take a
triangular prism with die same base area and altitude as shown in
Figure 1523b, Imagine two planes cutting the prism and dividing it
into three triangular pyramids as shown in Figure 1524, (Name the
two cutting planes that divide the prism as shown in Figure 1524,)
D.
V — =»*
w
Figure 1523
'51
(bj
15,5 Volumu of Pyramids and Cones 689
Figure 13S4
Pyramid (c) in Figure 1524 has the same base area and the same alti
tude as the given pyramid in Figure 1523, Therefore, by Theorem
15.12, they have the same volume. In Figure 15»23b, D£\\AC,
AD  C£, and ADEC is a parallelogram with AE one of its diagonals.
Therefore A, f>, E, C are coplanar and AADE s A CCA. Why? Think
of pyramids (a) and (b) in Figure 1524 as having bases ADE and ECA,
respectively, arid common vertex W. Then pyramids (a) and (b) have
the same base area (Why?) and the same altitude (the distance from
W lo plane A DEC). By Theorem 15.12. these two pyramids have the
same volume.
Next, consider WECB in Figure 1523. We have WE \ EC,
WB  UB. so WECB is a parallelograin with WC one of its diagonals.
Therefore W t E t C t B are coplanar and A WEC a ACBW. Think of
pyramids (b) and (c) in Figure 1524 as having bases WEC and CBW,
respectively, and common vertex A, Then pyramids (b) and (c) have
the same base area and the same altitude and, by Theorem 15.12, these
two pyramids have the same volume. Therefore all three pyramids,
(a). (b) t (c) in Figure 1524, have the same volume, say V, and the vol
ume of the prism in Figure 1523 is 3 V.
Xow consider ABC as the base of the prism in Figure 1523. Let
the area of AABC be S and let h be the altitude of the prism. Then
3V = Sh and V = $Sh.
But the given pyramid in Figure 1523 has the same base area S and
the same altitude h. Therefore the volume V of a triangular pyramid is
given by the formula
V= l 3 $h,
where S is the area of its base and h is its altitude. This result is out next
theort tii.
690 Areas and Volumes of Solids
Chapter 15
THEOREM 15.13 The volume of a triangular pyramid is one
third the product of its base area and its altitude.
The formula V = ^Sh holds for any pyramid as our next theorem
states.
THEOREM 15.14 The volume of a pyramid is onethird die
product of its base area and its altitude.
Let a pyramid with base area S and altitude h be given as
shown on the left in Figure 1525. Consider a triangular pyramid
having the same base area t the same altitude, and with Its base and the
base of the given pyramid in the same plane. It follows from the Pyra
mid Cross Section Theorem that cross sections of these two pyramids
formed by the same plane have the same area. By Cavalieri's Principle,
the two pyramids have the same volume. Since the volume of the tri
angular pyramid is yS/i, the volume of die given pyramid is also \Sh
and the proof is complete.
Figure 1525
Example 1 The dimensions of the base of a rectangular pyramid are
7 cm, by 11 cm. and its altitude is 16 cm. Find its volume,
Solution; The volume V* of the pyramid is given by the fonmila
V m \Sh. where S is the area of the base and h is Ihe altitude. We have
S = 77 and V = ^ • 77 • 10 = 410.
Therefore the volume is 4 10} cu. cm.
Figure 1526 shows a picture of a circular cone, Just as the defini
tion of a circular cylinder is analogous to the definition of a prism, the
definition of a circular cone is analogous to the definition of a pyramid.
15,5 Volume* of PynmWi and Cones 691
V
Figure 1526
Many of the terms used in defining a pyramid apply equally as well to a
circular cone. The base of a pyramid is a polygonal region. The base of
a circular cone is a circular region. Make appropriate changes in the
wording of the definition of a pyramid (Definition 1 5.7), refer to Figure
1526, and define a circular cone.
There are other cones in addition to circular cones, that is, cones
whose bases are not circular regions; but wc consider only circular
cones here. Therefore, when we speak of a cone, we mean a circular
cone. Also, when we speak of the base circle of a cone, we mean the
circle which is the boundary of the base.
Definition 15.8 (See Figure 1527.) If the center of the base
circle of a cone is the foot of the perpendicular from the ver
tex V to plane a, the cone is called a right circular cone.
Figure 1SS7
The following theorem is analogous to Theorems 15,4 and 15.5 on
pyramids and can be proved in a similar way, We give only an outline
of the proof and omit the details.
692
Af*« and Volumes of Solids
Chapter 15
THEOREM 15.15 (The Cone Cross Section Theorem) A cross
section of a cone of altitude h, made by a plane at a distance k from
the vertex, is a circular region whose area and the area r>f the base
are proportional to fc 2 and h 2 <
Outline of Proof: (See Figure 1528,) Let R be the base of the given
cone and let ft' be any cross section of the cone. Let C be the center of
the base circle, let P be any point on that circle, and let C and P be the
corresponding points in ft'. Ijet VA = h, VB = K CF = f, and
CF^S.
Figure 1528
2.
Then
AVU8~~ &VCA
Why?
(VC t k) = (VC t h)
Why?
AVCF* AVCP
Why?
(VC^jlVCr)
Why?
(VC,k,/) = {VC,h,f)
Why?
(k,/) = (h,r), ft*.
?{Kh),
/ =
T"
3, Since PC has a constant value regardless of the position of the
point F on the base circle, then FC has a constant value. Thus
all points F lie on a circle with radius ¥■ and center C". The cor
responding circular region is the cross section.
4. Let S' be the area of the circular cross section and let S be the
area of the base. Then
(S\ S) = [mV) 2 , **l f [{?)*, f*) = (*?, V) and (ff, S) = (**, k*).
This completes the outline of the proof.
15.5 Volumes of Pyramids and Cones 693
Cavalicri's Principle is used in the proof of the following theorem
which tells us how to find the volume of a cone,
THEOREM J 5. J 6 The volume of a circular cone is onethird the
product of die area of die base and the altitude.
Proof: The proof is similar to that of Theorem 15.14 and is left as an
exercise.
Figure 1529 shows a picture of a right circular cone. C is the center
of the base circle and P is a point of that circle. We call the distance
VP, which is the same for any point Pon the base circle, the slant height
of the cone and denote it by s.
Figure lS2»
If you imagine slitting the right cone of Fig^ire 1529 along VP and
unrolling its lateral surface onto a plane, you get the sector of a circle
shown on the right in Figure 1529, You learned in Chapter 14 that the
area of a sector of a circle is onehalf the product of the radius of the
sector and the length of the intercepted arc. In the case of a sector ob
tained by "unrolling" a cone, the radius of the sector is the slant height
of the cone, and the length of the intercepted arc is the circumference
of the base circle of the cone. Therefore a formula for the lateral sur
face area of a right circular cone is $sC f where $ is the slant height of
the cone and C is the circumference of the l»ase circle. If r Is the base
radius, then
%$C = 2$(2nr) 5= ms.
If S is the total surface area of a right circular cone, s its slant height,
and r the radius of its base circle, then a formula for the total surface
area is
S = J • * ■ 2<rrr + irr 2 or
This brings us to our next theorem.
S = <rrf(s + r).
694
Artat and Volumes of Solids
Chapter 15
THEOREM 15, 1 7 The lateral surface area of a right circular cone
of slant height s, base radius r, and base circumference C is f»0, or
ma, and its total surface area is «w(s + r)»
Example 2 TTie slant height of a right circular cone is 12 and the
radius of its base circle is 5. Find the lateral surface area and the total
surface area of the cone.
Solution: The lateral surface area is JsC, where
s = 12 and C = 2irr = 2n'B = lOir.
Therefore the lateral surface area is
^ 12 • l(fcr = 60sr.
The total surface area S is given by the formula
S = m{s + r).
Therefore
S = w*5(12 + 5} = 85?:,
EXERCISES 15,5
1. The length of one edge of the base of a regular triangular pyramid is
12 in. and the altitude of the pyramid is 18 in. Find the lateral surface
area of the pyramid. Find the total surface area. Find the volume. (Re
call that a regular pyramid is one whose base boundary is a regular poly
gon and that die eenter of the polygon is the foot of the perpendicular
from the vertex of the pyramid,)
Exercises 29 refer to the regular hexagonal pyramid in Figure 1530, with
VC = 1ft in., AB = 12 in., and P die projection of V on AB.
2. Find CP,
3. Find VP,
4. Find the area of die lateral face A V7J,
5. Find the lateral surface area
of the pyramid.
6. Find the area of A ABC.
7. Find the area of the base
of the pyramid.
8. Find the total surface area
of the pyramid.
9. Find the volume of the pyramid.
Figure 1530
15.5 Volumt* of Pyramid* and Cones 695
10. A plane bisects the altitude of a pyramid and is parallel to the base of
i he pyramid. Find the ratio of the volume of die pyramid above the
bisecting plane to the volume of the given pyramid (Hint: To what
numbers are the area of the cross section in the bisecting plane and the
area of the base proportional?)
11. Find the total surfoo© area and die volume of the right circular cone
shown in the figure*
12, Find the capacity in gallons of a right conical tank if it is 60 in* deep and
if the radius of its circular top is 28 in (To "see" the rank, invert the
cone of Exercise 11. Use *r = * 7 *.)
13. In the figure a right circular cone stands inside a right circular cylinder
of same base and altitude. Write a formula for the volume of that por
tion of the cylinder not occupied by the cone.
14, Figure 1531 shows two right cylinders having the same base area and
the same altitude. Figure 153 lb shows two cones with a common ver
tex V inside the cylinder. If V is midway between the bases of the
cylinder, show that the sum of the volumes of the two cones shown in
Figure 1331 b is equal to the volume of the cone shown in (a).
(a)
Figure 1531
696
/ >■ ,i ,iiii 'j'n lurries ot Solids
Chapter 15
15. The volume of the cone shown in the figure is 250 en. in. and its alti
tude is 15 in. A second tone is cut from the first by a plane parallel to
the base and 6 in. from the vertex. Find the volume of the smaller cone.
(Hint: If S and S* are the areas of the bases of the larger and smaller
cones, respectively, to what numbers are S r and S proportional?)
15 in.
Hi. challenge problem. If a plane parallel to the base of a cone (or
pyramid) cuts off anodicr cone (or pyramid), then the solid between
the cutting plane and the base is called a frustum. Figure 1 532b shows
a frustum of a right eone. If the radius of the base circle is 9 in., the
radius of the top circle 6 in., and the height of the frustum 12 in., find
the volume of the fnistum. (Hint: First find x t where x is the altitude of
the smaller cone in Figure 1532a.)
Figure 1532
(«)
17. challenge problem. The figure (on page 697) shows a picture of a
regular square pyramid, (Sec Exercise 1.) A plane parallel to the base
intersects the altitude of the pyramid at a point whose distance from the
vertex is onethird the distance from the vertex to the base. If the alti
tude of the pyramid is 24 in. and the length of an edge of the base is
18 in., find the lateral surface urea and the volume of the frustum. (See
Exercise 16.)
15.6 Sphtrt* and Sph«rie*I R«glons 697
15.6 SURFACE AREAS OF SPHERES AND VOLUMES OF
SPHERICAL REGIONS
The surface area of a sphere may be found very roughly by winding
a string around a hemisphere and by covering with string a circular
disk having the same radius as the sphere as suggested in Figure 1533.
Kifurc 1533
A comparison of the lengths of the two strings suggests that the surface
area of the hemisphere is twice the area of the circle. Since the area of a
circle is jrr 2 , this suggests that the surface area of a hemisphere is 2irr a
and that the surface area of a sphere is 4^.
We now proceed to a more sophisticated approach for finding the
surface area of a sphere, but first we need a definition.
Definition 15.9 A spherical region is the union of a sphere
and its interior.
Areas and Volumes of Solids
Chapter 15
Think of slicing a spherical region into n thin slices of thickness t
where nt is the diameter 2r of the sphere. (See Figure 1534.) This par
titions the surface into n zones.
Figure 1534
Figure 1535 suggests a vertical cross section C of the sphere made
by a plane through the center E of the sphere. The sphere has been
sliced by equally spaced horizontal planes into n zones.
Ji
■
 ■ ,
1 r
V'.
,'" R D
Ts
B7
T;
Figure 1535
One of these zones, Tg, for example, is the surface generated when
the arc AB rotates about the line FE. Let B' be the point in which the
tangent line to C at A intersects GB* Then AB is approximately equal
to AB and the area of the surface generated when AB' rotates about
FE is approximately equal to the area of the surface generated when
AB rotates about
15.6 Spheres and Spherical Regions 699
We have AB' 1AE. Why? Let D be the point on HE' such that
DA ± FK. Then AD X B'D. Why? Therefore &AB'D — AAEF.
(Show this by finding two pairs of congruent corresponding angles in
the two triangles.} Let s' = AB\ f a AD, r = AE, and x = AF. Then
(/* f to *)
and
s'* = rt.
Think of the zone T 2 as a narrow ribbon of width $> s is the length of the
are AB and is approximately equal to s\ Then the area of the zone is
approximately equal to the length of the ribbon, about 2mar, times the
width of the ribbon & Therefore the area of the zone is approximately
2ira»\ Now, 2tt.ts is approximately equal to 2ffxs'» and
Znx$ f a 2vrt
since s*i = rf . If we combine the areas of the n zones, we find the area
S of the sphere is given approximately by
S = n(2irrt)
= <2*rr)(nr)
= (2fTf)(2r)
"Ine total error introduced by using areas of ribbons to approximate
areas of zones can be made as small as desired if the thickness of the
slices is made small enough. The formula
S = 4wf*
for die surface area of a sphere is an exact formula, Our approach has
involved approximations, but our result is the correct one. In higher
mathematics, the area of a sphere is carefully defined and the assertion
that S = 4sr& is proved. We state it as a theorem,
THEOREM 15.18 The surface area of a sphere is 4tt times the
square of the radius of the sphere; that is, S = 4nr 2 .
In our discussion concerning Figure 1535* we showed that the area
Z of a zone T of a sphere is approximately equal to 2irrt , where r is the
radius of the sphere and t is the thickness of the zone. Actually,
Z = 2ffrf
is also an exact formula.
700
Areas and Volumn of Solids
Chapter 15
The formula for the volume of a sphere can now be obtained with
out difficulty. Xote that when we say "volume of a sphere/' we mean
"volume of a spherical region."
Suppose that the surface of the sphere is divided into n pieces of
equal areas by 'latitude" circles and "longitude" circles, regular spaced
as suggested in Figure 1536. Suppose that the pieces arc denoted by
figure Uk*
s±. s 2 ,... t s„. For i = 1, 2 n f let P ( be a point of s t . That is, P T is a
point of su P 2 is a point of *2, etc. Join each of the boundary points of
Si to the center C of the sphere with a segment. The union of these
segments and the piece $t encloses a portion of the spherical region
which is approximately a pyramid whose altitude is the radius of the
sphere and whose base is the piece sj. (Note that we could "flatten"
tine base by using a portion of the plane that is tangent to the sphere at
P\. Tlie error introduced by this flattening of the base when accumu
lated for all n pieces st, can be made as small as desired by making n
sufficientiy large,)
Recall that die volume of a pyramid is ^Sh t where S is the area of
the base and h is die altitude. Therefore die volume of one of the pyra
midal portions of the spherical region is approximately ]rS where r is
the radius of the sphere and 5 is the area of each .% By summing the
volumes of the portions of the sphere corresponding to all of the pieces
sj» we find that the volume V of the sphere is given approximately by
V = n(%rS)
= (W(«5)
= flrXV)
= $*rr3.
15.6 Spheres and Spherical Regions 701
Although we used approximations to obtain the formula for the
volume of a sphere, the formula
V =
is an exact formula. We write it formally as our last theorem.
THEOB EM 15. 19 If r is the radius of a sphere, the volume of the
sphere is jw 3 ,
EXERCISES 15.6
In working the exercises in this set, do not use a replacement for sr un
less instructed to do so.
1. Find the surface area and the volume of a sphere whose diameter is 12.
2. A sphere having a diameter of 14 in. is placed in a cubical box. If the
length of each edge of the box is 14 in., how many cubic inches of the
volume of the box are not occupied by the sphere? (Use n = ^f.)
3. The radii of two spheres are 3 in. and 6 in., respectively. Find the ratio
of their surface areas. Find the ratio of their volumes,
4. One sphere has a diameter that is three times that of a second sphere
Find the ratio of their surface areas. Find the ratio of their volumes.
5. In Exercise 4, if the larger sphere has a volume of 38,808, what is the
volume of the smaller sphere? Use v = If and find the radius of each
of the two spheres.
6. Given a hemisphere, a cylinder, and a cone such that the radii of the
base circles of the cylinder and the cone are equal to the radius of the
hemisphere. If the cylinder and the cone have altitudes equal to the
radius of the hemisphere, prove that the volume of the cylinder is equal
to the sum of the volumes of the hemisphere and the cone.
7. If the altitude of a circular cylinder is equal to the diameter of a sphere,
and if the radius of the base circle of the cylinder is equal to the radius of
the sphere, prove that the volume of the sphere is two thirds the volume
of the cylinder.
8. A sphere with radius 3^ in. is divided into 14 equal sections by planes
containing the same diameter of the sphere. Find the volume of each
section. {Use w = *jf%)
9. What is the largest radius a sphere eon have if the numerical value of
the surface area of the sphere is to be greater than or equal to the nu
merical value of the volume of the sphere?
10. The volumes of a sphere and a circular cone are equal, and the radius of
the sphere equals the radius of the base circle of the cone. Find the alti
tude of the oone in terms of die radius of the sphere.
702 Areas, and Volumes of Solids Chapter 15
11. The radius of a sphere is 5£. The volume of the sphere is equal to the
volume of a circular cone, The radius of the base circle of the cone is 5^.
Find the altitude of the cone.
12. The surfaces of Earth and its moon are approximately spheres with the
diameter of the moon about onefourth that of Earth, Find the surface
area (to the nearest 10,000 square miles) of the moon. {Use it = %fi
and 4(XK) miles as the radius of Earth.)
13. A water reservoir in the shape of a hemisphere has a diameter of 42 ft. If
1 cu. ft. of water weighs approximately 62.5 lb., how many tons (to the
nearest ton) of water docs die reservoir hold?
14. cm allence PROBLEM A sphere and a right circular cylinder have equal
volumes. The radius of the sphere is equal to the radius of the base circle
of the cylinder. Which has the larger surface area, the sphere or the
cylinder? Find the ratio of the larger to the smaller surface area.
CHAPTER SUMMARY
In this chapter we defined the following geometric solids. Review these
definitions.
PRISM RIGHT CIRCULAB CONE
RIGHT PRISM SPHERICAL REGION
PYRAMID PARALLELEPIPED
CIRCULAR CYLINDER RECTANGULAR
RIGHT CIRCULAR CYLINDER PARALLELEPIPED
CIRCULAR CONE CUBE
A pyramid is a REGULAR PYRAMID if its base boundary is a regular
polygon and if the foot of the perpendicular from its vertex to its base is die
tenter of the polygon. If a plane parallel to the plane of the base of a solid
intersects the solid, we defined the intersection to be a CROSS SECTION
of the solid provided the intersection consists of more than a single point
We proved that all cross sections of a prism (cylinder) have the same
area and that the area of a cross section of a pyramid (cone) and the area
of the base are proportional to k 2 a